Get summaries, questions, answers, solutions, notes, extras, workbook solutions, PDF and guide of chapter 11 Calorimetry: ICSE Class 10 Physics which is part of the syllabus of students studying under the Council for the Indian School Certificate Examinations board. These solutions, however, should only be treated as references and can be modified/changed.
Summary
Heat is a form of energy that flows from a hotter object to a cooler object when they are in contact. Every substance is made of tiny particles called molecules, which are always moving and attracting each other. These molecules have internal energy due to their motion (kinetic energy) and attraction (potential energy). The total internal energy of all molecules in a substance is its thermal energy.
Temperature tells us how hot or cold something is. It also determines the direction in which heat will flow. The S.I. unit of heat is the joule (J). Another common unit is the calorie (cal), defined as the heat needed to raise the temperature of 1 gram of water by 1 degree Celsius. Temperature is measured in degrees Celsius (°C) or Kelvin (K).
Different objects require different amounts of heat to change their temperature by the same amount. Heat capacity is the amount of heat needed to raise an object’s temperature by 1°C (or 1K). Specific heat capacity is the heat needed to raise the temperature of 1 gram (or 1 kilogram) of a substance by 1°C (or 1K). Water has a high specific heat capacity, meaning it absorbs a lot of heat for a small temperature rise and releases a lot of heat when it cools. This property makes water useful in car radiators and hot water bottles. The climate near large bodies of water is often moderate because water heats up and cools down slowly.
Calorimetry is the measurement of heat. A calorimeter is a device used for this, often a copper vessel. The principle of mixtures states that when a hot body and a cold body are mixed in an insulated system, the heat lost by the hot body equals the heat gained by the cold body until they reach the same temperature.
Substances can exist in different states: solid, liquid, or gas. Changing from one state to another is a change of phase. This change happens at a constant temperature. For example, ice (solid) melts into water (liquid) at 0°C. Water boils into steam (gas) at 100°C. The heat absorbed or released during a phase change without a temperature change is called latent heat. The specific latent heat of fusion is the heat needed to change a unit mass of a substance from solid to liquid at its melting point. For ice, this is why it effectively cools drinks; it absorbs heat from the drink to melt. Similarly, specific latent heat of vaporization is for the liquid-to-gas change.
A heating curve is a graph that shows how the temperature of a substance changes as it is heated over time. During phase changes, the temperature remains constant on the graph, forming flat sections. The melting point of a substance can be affected by pressure and impurities. For example, ice melts at a lower temperature under increased pressure. Adding impurities like salt to ice also lowers its melting point, which is used in making freezing mixtures.
Workbok Solutions(Concise/Selina)
Exercise (A)
MCQ
1. The total internal energy of all molecules of a substance is called:
(a) magnetic energy
(b) heat energy
(c) thermal energy
(d) electrical energy
Answer: (c) thermal energy
2. The unit of heat is:
(a) Watt
(b) Joule
(c) Calorie
(d) Both (b) and (c)
Answer: (d) Both (b) and (c)
3. The specific heat capacity of water is :
(a) 4200 J kg⁻¹ K⁻¹
(b) 420 J g⁻¹ K⁻¹
(c) 0.42 J g⁻¹ K⁻¹
(d) 4.2 J kg⁻¹ K⁻¹
Answer: (a) 4200 J kg⁻¹ K⁻¹
4. Temperature determines the direction of :
(a) flow of heat
(b) flow of energy
(c) flow of motion
(d) none of the above
Answer: (a) flow of heat
5. The amount of heat energy contained in a body depends on its :
(a) mass
(b) temperature
(c) material of the body
(d) all of the above
Answer: (d) all of the above
6. The correct relation between heat capacity and specific heat capacity is :
(a) C’ = m x c
(b) C’ = 1/mc
(c) C’ = m/c
(d) C = mc’
Answer: (a) C’ = m x c
7. A good conductor of heat has a ……. specific heat capacity, while a bad conductor has a ……. specific heat capacity.
(a) low, high
(b) low, low
(c) high, low
(d) high, high
Answer: (a) low, high
8. The specific heat capacity is maximum for :
(a) copper
(b) zinc
(c) iron
(d) hydrogen
Answer: (d) hydrogen
9. The principle of calorimetry is based on :
(a) law of conservation of heat
(b) law of conservation of energy
(c) both (a) and (b)
(d) conservation of momentum
Answer: (c) both (a) and (b)
10. Heat is measured by :
(a) thermometer
(b) barometer
(c) principle of calorimetry
(d) both (a) and (b)
Answer: (c) principle of calorimetry
11. Assertion (A): The specific heat capacity of a substance is the amount of heat required to raise the temperature of unit mass of that substance by 1°C. Reason (R) : The specific heat capacity of a substance is not its characteristic property.
(a) both A and R are true and R is the correct explanation of A
(b) both A and R are True and R is not the correct explanation of A
(c) assertion is false but reason is true
(d) assertion is true but reason is false
Answer: (d) assertion is true but reason is false
12. Assertion (A): The principle of the method of mixtures involves mixing substances at different temperatures to find the final temperature. Reason (R) : The law of conservation of energy states that the energy is neither created nor destroyed in an isolated system.
(a) both A and R are true and R is the correct explanation of A
(b) both A and R are True and R is not the correct explanation of A
(c) assertion is false but reason is true
(d) assertion is true but reason is false
Answer: (a) both A and R are true and R is the correct explanation of A
Very Short Answer Type Questions
1. Name the S.I. unit of heat.
Answer: The S.I. unit of heat is joule (symbol J).
2. How is heat capacity of a body related to specific heat capacity of its substance ?
Answer: The heat capacity C’ of a body is related to the specific heat capacity c of its substance by the relation: Heat capacity C’ = mass m × specific heat capacity c.
3. Name a liquid which has the highest specific heat capacity.
Answer: Water has an unusually high specific heat capacity. The specific heat capacity is maximum for hydrogen, but among common liquids, water has a very high specific heat capacity.
4. Write the approximate value of specific heat capacity of water in S.I. unit.
Answer: The approximate value of specific heat capacity of water is 4200 J kg⁻¹ K⁻¹. More precisely, the specific heat capacity of water is 4180 J kg⁻¹ K⁻¹, but for convenience, its value is often taken as 4200 J kg⁻¹ K⁻¹.
5. Write the expression for the heat energy Q received by m kg of a substance of specific heat capacity c J kg⁻¹ K⁻¹ when it is heated through ∆t °C.
Answer: The expression for the heat energy Q received by m kg of a substance of specific heat capacity c J kg⁻¹ K⁻¹ when it is heated through ∆t °C (or ∆t K) is Q = m × c × ∆t joule.
6. Same amount of heat is supplied to two liquids A and B. The liquid A shows a greater rise in temperature. What can you say about the heat capacity of A as compared to that of B ?
Answer: If liquid A shows a greater rise in temperature for the same amount of heat supplied, it means that the heat capacity of A is less than that of B.
7. Give one example each where high specific heat capacity of water is used (i) as coolant, (ii) as heat reservoir.
Answer: (i) Water is used as an effective coolant in radiators in cars and generators.
(ii) In cold countries, water is used as a heat reservoir for wine and juice bottles to avoid their freezing.
8. A liquid X has specific heat capacity higher than the liquid Y. Which liquid is useful as (i) coolant in car radiators, and (ii) heat reservior to keep juice bottles without freezing ?
Answer: (i) Liquid X is useful as a coolant in car radiators.
(ii) Liquid X is useful as a heat reservoir to keep juice bottles without freezing.
Short Answer Type Questions
1. Define the term heat.
Answer: Heat is that form of energy which flows from a hot body to a cold body when they are kept in contact.
2. Define the term calorie. How is it related to joule ?
Answer: One calorie is the quantity of heat energy required to raise the temperature of 1 g of water through 1°C. The precise definition of calorie (which is also called 15°C calorie) is the heat energy required to raise the temperature of 1 g of water from 14.5°C to 15.5°C.
The unit calorie is related to the S.I. unit joule as follows: 1 calorie (or 1 cal) = 4.186 J or 4.2 J nearly. For calculations, we generally take 1 cal = 4.2 J.
3. Define one kilo-calorie of heat.
Answer: One kilo-calorie is the heat energy required to raise the temperature of 1 kg of water from 14.5°C to 15.5°C. Also, 1 kilo-calorie = 1000 calorie = 4200 J nearly.
4. Define temperature and name its S.I. unit.
Answer: Temperature is a parameter which tells the thermal state of a body (i.e., the degree of hotness or coldness of the body). It determines the direction of flow of heat when two bodies at different temperatures are placed in contact. The S.I. unit of temperature is kelvin (symbol K).
5. State three differences between heat and temperature.
Answer: Three differences between heat and temperature are:
- Heat is that form of energy which flows from a hot body to a cold body when they are kept in contact, while temperature is a parameter which determines the direction of flow of heat on keeping the two bodies at different temperatures in contact.
- The S.I. unit of heat is joule (J), while the S.I. unit of temperature is kelvin (K).
- The amount of heat contained in a body depends on mass, temperature and substance of body, while the temperature of a body depends on the average kinetic energy of its molecules due to their random motion.
6. State the principle of calorimetry.
Answer: The principle of calorimetry, also known as the principle of method of mixtures, states that when a hot body is mixed (or is kept in contact) with a cold body, heat energy passes from the hot body to the cold body, till both the bodies attain the same temperature. If no heat energy is lost to the surroundings (i.e., if the system is perfectly insulated), then Heat energy lost by the hot body = Heat energy gained by the cold body. This principle is based on the law of conservation of energy.
7. Define the term heat capacity and state its S.I. unit.
Answer: The heat capacity of a body is the amount of heat energy required to raise its temperature by 1°C (or 1 K). The S.I. unit of heat capacity is joule per kelvin (or J K⁻¹). It is also written as joule per degree C (or J °C⁻¹), but since S.I. unit of temperature is kelvin (K), it is more appropriate to write the S.I. unit of heat capacity as J K⁻¹ instead of J °C⁻¹.
8. Define the term specific heat capacity and state its S.I. unit.
Answer: The specific heat capacity of a substance is the amount of heat energy required to raise the temperature of unit mass of that substance through 1°C (or 1 K). The S.I. unit of specific heat capacity is joule per kilogram per kelvin (or J kg⁻¹ K⁻¹) or joule per kilogram per degree celsius (or J kg⁻¹ °C⁻¹).
9. State three differences between the heat capacity and specific heat capacity.
Answer: Three differences between the heat capacity and specific heat capacity are:
- Heat capacity is the amount of heat energy required to raise the temperature of the entire body by 1°C, while specific heat capacity is the amount of heat energy required to raise the temperature of unit mass of the body by 1°C.
- Heat capacity depends both on the nature of the substance and mass of the body. More the mass of the body, more is its heat capacity. Specific heat capacity does not depend on the mass of the body, but it is the characteristic property of the substance of the body.
- Heat capacity C’ = Q/∆t = mass m × specific heat capacity c, while specific heat capacity c = Q/(m∆t) = Heat capacity C’/mass m.
- The unit of heat capacity is J K⁻¹, while the unit of specific heat capacity is J kg⁻¹ K⁻¹.
10. What do you mean by the following statements :
(i) the heat capacity of a body is 50 J K⁻¹?
(ii) the specific heat capacity of copper is 0.4 J g⁻¹ K⁻¹?
Answer: (i) If the heat capacity of a body is 50 J K⁻¹, it means that 50 J heat energy is required to raise the temperature of that body by 1 K (or 1°C).
(ii) If the specific heat capacity of copper is 0.4 J g⁻¹ K⁻¹, it means that the heat energy required to raise the temperature of 1 g of copper by 1 K (or 1°C) is 0.4 J.
11. Specific heat capacity of a substance A is 3.8 J g⁻¹ K⁻¹ and of the substance B is 0.4 J g⁻¹ K⁻¹. Which substance is a good conductor of heat ? How did you arrive at your conclusion ?
Answer: Substance B is a good conductor of heat. Usually a good conductor of heat has a low specific heat capacity, while a bad conductor has a high specific heat capacity. Since substance B has a lower specific heat capacity (0.4 J g⁻¹ K⁻¹) compared to substance A (3.8 J g⁻¹ K⁻¹), B is a better conductor of heat. The substance with low specific heat capacity shows a rapid and high rise in temperature for the same heat supply and mass, indicating it conducts heat better.
12. Name two factors on which the heat energy librated by a body on cooling depends.
Answer: The quantity of heat energy liberated by a body on cooling depends on three factors:
(i) Mass of the body.
(ii) The fall in the temperature of the body.
(iii) The material or substance of the body (i.e., its specific heat capacity).
So, two factors are the mass of the body and the fall in its temperature.
13. Name three factors on which the heat energy absorbed by a body depends and state how does it depend on them.
Answer: The quantity of heat energy absorbed to increase the temperature of a body depends on three factors:
(1) The mass of the body (m): The amount of heat energy absorbed (Q) is directly proportional to the mass of the object, i.e., Q ∝ m.
(2) The increase in temperature of the body (∆t): The amount of heat energy absorbed (Q) is directly proportional to the rise in temperature ∆t, i.e., Q ∝ ∆t.
(3) The material (or substance) of the body: This characteristic is expressed in terms of its specific heat capacity (c). The amount of heat energy absorbed (Q) is proportional to c, i.e., Q ∝ c.
Thus, Q = c m ∆t.
14. Two blocks P and Q of different metals having their mass in the ratio 2 : 1 are given same amount of heat. Their temperature rises by same amount. Compare their specific heat capacities.
Given:
Let the mass, specific heat capacity, heat supplied, and rise in temperature for block P be m_P, c_P, Q_P, and ΔT_P respectively.
Let the mass, specific heat capacity, heat supplied, and rise in temperature for block Q be m_Q, c_Q, Q_Q, and ΔT_Q respectively.
According to the question:
Ratio of masses, m_P : m_Q = 2 : 1
Amount of heat supplied is the same, Q_P = Q_Q
Rise in temperature is the same, ΔT_P = ΔT_Q
To find:
The ratio of their specific heat capacities, c_P : c_Q = ?
Solution:
We know that the amount of heat (Q) absorbed by a body is given by the formula:
Q = m * c * ΔT
where,
m = mass of the body
c = specific heat capacity of the body
ΔT = rise in temperature
For block P:
Q_P = m_P * c_P * ΔT_P —(i)
For block Q:
Q_Q = m_Q * c_Q * ΔT_Q —(ii)
Since the amount of heat supplied and the rise in temperature are the same for both blocks (Q_P = Q_Q and ΔT_P = ΔT_Q), we can equate equations (i) and (ii):
m_P * c_P * ΔT_P = m_Q * c_Q * ΔT_Q
Cancelling the common term ΔT (since ΔT_P = ΔT_Q) from both sides, we get:
m_P * c_P = m_Q * c_Q
We are given that the ratio of masses m_P : m_Q = 2 : 1, or m_P / m_Q = 2 / 1.
Rearranging the equation to find the ratio of specific heat capacities (c_P / c_Q):
c_P / c_Q = m_Q / m_P
Now, substituting the ratio of masses:
c_P / c_Q = 1 / 2
Therefore, the ratio of the specific heat capacities of block P to block Q is 1 : 2.
15. What is the principle of method of mixture ? What other name is given to it? Name the law on which this principle is based.
Answer: The principle of method of mixture states that when a hot body is mixed (or is kept in contact) with a cold body, heat energy passes from the hot body to the cold body, till both the bodies attain the same temperature. If no heat energy is lost to the surroundings, then heat energy lost by the hot body is equal to heat energy gained by the cold body.
Another name given to it is the principle of calorimetry.
This principle is based on the law of conservation of energy.
16. Why do the farmers fill their fields with water on a cold winter night?
Answer: Farmers fill their fields with water to protect the crops from frost. On a cold winter night, when the atmospheric temperature falls below 0°C, water in the fine capillaries of plants freezes, so the veins burst due to the increase in volume of water on freezing. As a result, plants die and the crop gets destroyed. In order to save crop on such cold nights, farmers fill their fields with water because water has a high specific heat capacity, so it does not allow the temperature in the surrounding area of plants to fall up to 0°C.
17. Water is used in hot water bottles for fomentation. Give a reason.
Answer: Hot water bottles are used for fomentation. The reason is that water does not cool quickly due to its high specific heat capacity, so a hot water bottle provides more heat energy for fomentation over a longer period.
18. What property of water makes it an effective coolant?
Answer: Water is used as an effective coolant because of its high specific heat capacity. Water in pipes can extract more heat from the surroundings without much rise in its temperature because of its high specific heat capacity.
19. Why is the base of a cooking pan made thick and heavy?
Answer: The base of a cooking pan is made thick. By making the base of the cooking pan thick, its heat capacity becomes large due to which it gets heated slowly and imparts sufficient heat energy at a slow rate to the food for its proper cooking. After cooking, it also keeps the food warm for a longer period. If it is also heavy, this contributes to a larger heat capacity.
Long Answer Type Questions
1. A mass m₁ of a substance of specific heat capacity c₁ at temperature t₁ is mixed with a mass m₂ of other substance of specific heat capacity c₂ at a lower temperature t₂. Deduce the expression for the temperature t of the mixture. State the assumption made, if any.
Answer: Let a substance A of mass m₁, specific heat capacity c₁ at a higher temperature t₁ be mixed with another substance B of mass m₂, specific heat capacity c₂ at a lower temperature t₂ (i.e., t₂ < t₁). If the final temperature of mixture after heat exchange becomes t, then
Fall in temperature of the substance A = t₁ – t
Rise in temperature of substance B = t – t₂
Heat energy lost by A = m₁ × c₁ × fall in temperature = m₁ c₁ (t₁ – t)
Heat energy gained by B = m₂ × c₂ × rise in temperature = m₂ c₂ (t – t₂)
If no heat energy is lost to the surroundings, then by the principle of method of mixtures,
Heat energy lost by A = Heat energy gained by B
or m₁ c₁ (t₁ – t) = m₂ c₂ (t – t₂)
m₁c₁t₁ – m₁c₁t = m₂c₂t – m₂c₂t₂
m₁c₁t₁ + m₂c₂t₂ = m₂c₂t + m₁c₁t
m₁c₁t₁ + m₂c₂t₂ = t (m₂c₂ + m₁c₁)
So, the temperature of the mixture t = (m₁c₁t₁ + m₂c₂t₂) / (m₁c₁ + m₂c₂).
The assumption made is that there is no loss of heat energy to the surroundings.
2. Discuss the role of high specific heat capacity of water with reference to climate in coastal areas.
Answer: The climate near the sea shore is moderate. The specific heat capacity of water is very high (= 1000 cal kg⁻¹ °C⁻¹ or 4200 J kg⁻¹ K⁻¹). It is about five times as high as that of sand. Hence, the heat energy required for the same rise in temperature by a certain mass of water is nearly five times more than that required by the same mass of sand. Similarly, a certain mass of water imparts nearly five times more heat energy than that given by the same mass of sand for the same fall in temperature. As such, sand (or earth) gets heated or cooled more rapidly as compared to water under similar conditions (exposure to the sun). Thus, near the sea shore, there develops a large difference in temperature between the land and sea surfaces due to which convection air currents are set up. The cold air blows from the land towards the sea during the night (i.e., land breeze) and during the day, cold air blows from the sea towards the land (i.e., sea breeze). These breezes make the climate near the sea shore moderate.
3. (a) What is a calorimeter ?
(b) Name the material of which it is made of. Give two reasons for using the material stated by you.
(c) Out of the three metals A, B and C of specific heat capacity 900 J kg⁻¹ °C⁻¹, 380 J kg⁻¹ °C⁻¹ and 460 J kg⁻¹ °C⁻¹ respectively, which will you prefer for calorimeter ? Give reason.
(d) How is the loss of heat due to radiation minimised in a calorimeter?
Answer: (a) A calorimeter is a cylindrical vessel which is used to measure the amount of heat gained (or lost) by a body when it is mixed with another body or substance.
(b) It is made up of a thin sheet of copper. Two reasons for using copper are: (i) copper is a good conductor of heat, so the vessel soon acquires the temperature of its contents, and (ii) copper has low specific heat capacity so the amount of heat energy taken by the calorimeter itself from the contents to acquire its temperature, is very small.
(c) Out of the three metals A (900 J kg⁻¹ °C⁻¹), B (380 J kg⁻¹ °C⁻¹), and C (460 J kg⁻¹ °C⁻¹), metal B with the specific heat capacity of 380 J kg⁻¹ °C⁻¹ will be preferred for the calorimeter. The reason is that a material with low specific heat capacity is chosen so that the calorimeter takes a negligible amount of heat from its contents to attain the temperature of the contents.
(d) The loss of heat due to radiation is minimised in a calorimeter because the outer and inner surfaces of the vessel are polished.
Numerical Questions
1. By imparting heat to a body, its temperature rises by 15°C. What is the corresponding rise in temperature on kelvin scale ?
Answer:
Given:
Rise in temperature on Celsius scale (ΔT_C) = 15°C
To find:
Corresponding rise in temperature on Kelvin scale (ΔT_K) = ?
Solution:
The relationship between the temperature on the Kelvin scale (T_K) and the temperature on the Celsius scale (T_C) is given by the formula:
T_K = T_C + 273.15
Let the initial temperature of the body be T1_C in Celsius and T1_K in Kelvin.
Let the final temperature of the body be T2_C in Celsius and T2_K in Kelvin.
The rise in temperature on the Celsius scale is:
ΔT_C = T2_C – T1_C
=> 15°C = T2_C – T1_C
The rise in temperature on the Kelvin scale is:
ΔT_K = T2_K – T1_K
Now, we can express the Kelvin temperatures in terms of Celsius temperatures using the conversion formula:
=> ΔT_K = (T2_C + 273.15) – (T1_C + 273.15)
=> ΔT_K = T2_C + 273.15 – T1_C – 273.15
=> ΔT_K = T2_C – T1_C
Since we know that T2_C – T1_C = ΔT_C = 15°C, we can substitute this value:
=> ΔT_K = 15 K
Therefore, a rise in temperature of 15°C is equal to a rise in temperature of 15 K.
2. (a) Calculate the heat capacity of a copper vessel of mass 200 g if the specific heat capacity of copper is 410 J kg⁻¹ K⁻¹.
(b) How much heat energy will be required to increase the temperature of the vessel in part (a) from 25°C to 35°C ?
Answer:
(a) To calculate the heat capacity of the copper vessel.
Given:
Mass of the copper vessel (m) = 200 g
Specific heat capacity of copper (c) = 410 J kg⁻¹ K⁻¹
To find:
Heat capacity of the vessel (C’) = ?
Solution:
First, we need to convert the mass from grams (g) to kilograms (kg) to match the units of the specific heat capacity.
1 kg = 1000 g
=> m = 200 / 1000 kg
=> m = 0.2 kg
The formula for heat capacity is:
Heat Capacity (C’) = mass (m) × specific heat capacity (c)
=> C’ = 0.2 kg × 410 J kg⁻¹ K⁻¹
=> C’ = 82 J K⁻¹
The heat capacity of the copper vessel is 82 J K⁻¹ (or 82 J/°C).
(b) To calculate the heat energy required.
Given:
Heat capacity of the vessel (C’) = 82 J K⁻¹ (from part a)
Initial temperature (T₁) = 25°C
Final temperature (T₂) = 35°C
To find:
Heat energy required (Q) = ?
Solution:
First, we calculate the change in temperature (ΔT).
ΔT = Final temperature – Initial temperature
=> ΔT = T₂ – T₁
=> ΔT = 35°C – 25°C
=> ΔT = 10°C (A change of 10°C is equal to a change of 10 K)
Now, we use the formula for heat energy:
Heat energy (Q) = Heat capacity (C’) × Change in temperature (ΔT)
=> Q = 82 J K⁻¹ × 10 K
=> Q = 820 J
Therefore, 820 J of heat energy will be required to increase the temperature of the vessel.
3. A piece of iron of mass 2.0 kg has a heat capacity of 966 J K⁻¹. Find : (i) heat energy needed to warm it by 15°C, and (ii) its specific heat capacity in S.I. unit.
(i) Find the heat energy needed to warm it by 15°C.
Answer:
Given:
Heat capacity (C’) = 966 J K⁻¹
Change in temperature (ΔT) = 15°C
Note: A change in temperature of 15°C is equal to a change in temperature of 15 K. So, ΔT = 15 K.
To find:
Heat energy (Q) = ?
Solution:
The relationship between heat energy, heat capacity, and change in temperature is given by the formula:
Q = C’ × ΔT
=> Q = 966 J K⁻¹ × 15 K
=> Q = 14490 J
Therefore, the heat energy needed to warm the piece of iron is 14490 J.
(ii) Find its specific heat capacity in S.I. unit.
Answer:
Given:
Mass of iron (m) = 2.0 kg
Heat capacity (C’) = 966 J K⁻¹
To find:
Specific heat capacity (c) = ?
Solution:
The relationship between heat capacity (C’), mass (m), and specific heat capacity (c) is given by the formula:
C’ = m × c
To find the specific heat capacity (c), we can rearrange the formula:
c = C’ / m
=> c = 966 J K⁻¹ / 2.0 kg
=> c = 483 J kg⁻¹ K⁻¹
Therefore, the specific heat capacity of iron is 483 J kg⁻¹ K⁻¹.
4. Calculate the amount of heat energy required to raise the temperature of 200 g of copper from 20°C to 70°C. Specific heat capacity of copper = 390 J kg⁻¹ K⁻¹.
Answer:
Solution:
First, we need to ensure all units are consistent with the specific heat capacity unit (J kg⁻¹ K⁻¹). We must convert the mass from grams (g) to kilograms (kg).
Mass (m) = 200 g
=> m = 200 / 1000 kg
=> m = 0.2 kg
Next, we calculate the change in temperature (ΔT).
ΔT = Final temperature – Initial temperature
=> ΔT = T₂ – T₁
=> ΔT = 70°C – 20°C
=> ΔT = 50°C
A change in temperature of 50°C is equivalent to a change in temperature of 50 K.
The formula for the heat energy (Q) required to change the temperature of a substance is:
Q = m × c × ΔT
Now, we substitute the values into the formula:
Q = 0.2 kg × 390 J kg⁻¹ K⁻¹ × 50 K
=> Q = 78 × 50 J
=> Q = 3900 J
Therefore, the amount of heat energy required to raise the temperature of the copper is 3900 J.
5. 1300 J of heat energy is supplied to raise the temperature of 0.5 kg of lead from 20°C to 40°C. Calculate the specific heat capacity of lead.
Answer:
Given:
Heat energy supplied (Q) = 1300 J
Mass of lead (m) = 0.5 kg
Initial temperature (T₁) = 20°C
Final temperature (T₂) = 40°C
To find:
Specific heat capacity of lead (c) = ?
Solution:
First, we need to calculate the change in temperature (ΔT).
ΔT = Final temperature – Initial temperature
=> ΔT = T₂ – T₁
=> ΔT = 40°C – 20°C
=> ΔT = 20°C
Now, we use the formula for heat energy:
Q = m × c × ΔT
To find the specific heat capacity (c), we can rearrange the formula:
c = Q / (m × ΔT)
Now, we substitute the given values into the formula:
=> c = 1300 / (0.5 × 20)
=> c = 1300 / 10
=> c = 130 J/kg°C
Therefore, the specific heat capacity of lead is 130 J/kg°C.
6. A car’s cooling system uses water to absorb heat from the engine. If 5 kg of water absorbs 420 kJ of heat, what is the temperature increase of water? (Specific heat capacity of water = 4200 J kg⁻¹K⁻¹).
Answer:
Given:
Mass of water (m) = 5 kg
Heat absorbed (Q) = 420 kJ
Specific heat capacity of water (c) = 4200 J kg⁻¹K⁻¹
To find:
Temperature increase of water (ΔT) = ?
Solution:
The relationship between heat energy, mass, specific heat capacity, and temperature change is given by the formula:
Q = m * c * ΔT
Where:
Q = Heat absorbed
m = Mass
c = Specific heat capacity
ΔT = Change in temperature
First, we must ensure all units are consistent. The heat absorbed (Q) is given in kilojoules (kJ), while the specific heat capacity (c) is in Joules (J). We need to convert kJ to J.
1 kJ = 1000 J
=> Q = 420 kJ = 420 × 1000 J
=> Q = 420,000 J
Now, we rearrange the formula to solve for the temperature increase (ΔT):
Q = m * c * ΔT
=> ΔT = Q / (m * c)
Substitute the given values into the formula:
=> ΔT = 420,000 / (5 * 4200)
=> ΔT = 420,000 / 21,000
=> ΔT = 20 K
Since a change of 1 Kelvin is equal to a change of 1 degree Celsius, the temperature increase is 20 K or 20 °C.
7. Find the time taken by a 500 W heater to raise the temperature of 50 kg of material of specific heat capacity 960 J kg⁻¹ K⁻¹ from 18°C to 38°C. Assume that all the heat energy supplied by heater is given to the material.
Answer:
To find:
Time taken (t) = ?
Solution:
First, we need to calculate the change in temperature (ΔT).
ΔT = Final temperature – Initial temperature
=> ΔT = 38°C – 18°C
=> ΔT = 20°C
A change in temperature of 20°C is equivalent to a change of 20 K.
So, ΔT = 20 K.
Next, we calculate the heat energy (Q) absorbed by the material using the formula:
Q = m * c * ΔT
=> Q = 50 kg * 960 J kg⁻¹ K⁻¹ * 20 K
=> Q = 960,000 J
The energy (E) supplied by the heater is given by the formula:
E = P * t
According to the question, the energy supplied by the heater is equal to the heat energy absorbed by the material.
Therefore, E = Q.
P * t = Q
=> 500 W * t = 960,000 J
Now, we can solve for the time (t).
t = Q / P
=> t = 960,000 J / 500 W
=> t = 1920 s
The time taken by the heater is 1920 seconds. To express this in minutes:
t = 1920 / 60
=> t = 32 minutes.
8. A fire truck uses 10,000 litres of water to extinguish a fire, cooling down the burning material from 500°C to 100°. If the temperature of water increases by 40°C, how much heat energy is absorbed by it? (Specific heat capacity of water = 4200 J kg⁻¹K⁻¹).
Answer:
To find:
Heat energy absorbed by water (Q) = ?
Solution:
First, we need to find the mass (m) of the water from its volume. The density of water is approximately 1 kg/litre .
Mass (m) = Volume (V) × Density (ρ)
=> m = 10,000 litres × 1 kg/litre
=> m = 10,000 kg
The formula for the heat energy absorbed is:
Q = mcΔT
Note: A change in temperature of 40°C is equal to a change in temperature of 40 K.
Substituting the values into the formula:
=> Q = (10,000 kg) × (4200 J kg⁻¹K⁻¹) × (40 K)
=> Q = 10,000 × 4200 × 40 J
=> Q = 42,000,000 × 40 J
=> Q = 1,680,000,000 J
The heat energy absorbed by the water is 1,680,000,000 Joules. This can also be expressed in megajoules (MJ).
=> Q = 1,680 MJ.
168 × 10⁷ J
9. An electric heater of power 600 W raises the temperature of 4.0 kg of a liquid from 10.0 °C to 15.0 °C in 100 s. Calculate : (i) the heat capacity of 4.0 kg of liquid, and (ii) the specific heat capacity of liquid.
(i) Calculate the heat capacity of 4.0 kg of liquid.
Answer:
Given:
Power of the heater (P) = 600 W
Mass of the liquid (m) = 4.0 kg
Initial temperature (T_initial) = 10.0 °C
Final temperature (T_final) = 15.0 °C
Time (t) = 100 s
To find:
The heat capacity (C’) of 4.0 kg of liquid = ?
Solution:
First, we calculate the heat energy (Q) supplied by the heater. We assume that all the electrical energy is converted into heat energy.
The formula for energy supplied is:
Energy (Q) = Power (P) × Time (t)
=> Q = 600 W × 100 s
=> Q = 60000 J
Next, we calculate the change in temperature (ΔT).
ΔT = Final temperature – Initial temperature
=> ΔT = 15.0 °C – 10.0 °C
=> ΔT = 5.0 °C
Now, we can calculate the heat capacity (C’). The formula for heat capacity is:
Heat capacity (C’) = Heat energy supplied (Q) / Change in temperature (ΔT)
=> C’ = 60000 J / 5.0 °C
=> C’ = 12000 J/°C
(ii) Calculate the specific heat capacity of liquid.
Answer:
Given:
Heat capacity (C’) = 12000 J/°C (from part i)
Mass of the liquid (m) = 4.0 kg
To find:
The specific heat capacity (c) of the liquid = ?
Solution:
The relationship between heat capacity (C’) and specific heat capacity (c) is given by the formula:
Heat capacity (C’) = Mass (m) × Specific heat capacity (c)
To find the specific heat capacity, we rearrange the formula:
c = C’ / m
=> c = 12000 J/°C / 4.0 kg
=> c = 3000 J/(kg °C)
10. 0.5 kg of lemon squash at 30°C is placed in a refrigerator which can remove heat at an average rate of 30 J s⁻¹. How long will it take to cool the lemon squash to 5°C ? Specific heat capacity of squash = 4200 J kg⁻¹ K⁻¹.
Answer:
Given:
Mass of lemon squash (m) = 0.5 kg
Initial temperature (T₁) = 30°C
Final temperature (T₂) = 5°C
Rate of heat removal (Power, P) = 30 J s⁻¹
Specific heat capacity of squash (c) = 4200 J kg⁻¹ K⁻¹
To find:
Time taken (t) = ?
Solution:
First, we need to calculate the total heat (Q) to be removed from the lemon squash to cool it from 30°C to 5°C.
The formula for the heat removed is:
Q = m * c * ΔT
Where,
ΔT = Change in temperature
=> ΔT = Initial Temperature – Final Temperature
=> ΔT = T₁ – T₂
=> ΔT = 30°C – 5°C
=> ΔT = 25°C
Note: A change in temperature of 25°C is equal to a change of 25 K.
Now, we can calculate the total heat (Q) to be removed:
Q = 0.5 * 4200 * 25
=> Q = 2100 * 25
=> Q = 52500 J
The refrigerator removes heat at a rate (P) of 30 Joules per second (30 J s⁻¹). The relationship between total heat removed (Q), rate of heat removal (P), and time taken (t) is:
P = Q / t
To find the time (t), we can rearrange the formula:
t = Q / P
Now, substitute the values of Q and P:
t = 52500 / 30
=> t = 1750 s
Therefore, it will take 1750 seconds to cool the lemon squash to 5°C.
11. A mass of 50 g of a certain metal at 150°C is immersed in 100 g of water at 11°C. The final temperature is 20°C. Calculate the specific heat capacity of the metal. Assume that the specific heat capacity of water is 4.2 J g⁻¹ K⁻¹.
Answer:
Given:
- For the metal:
- Mass (m_metal) = 50 g
- Initial temperature (T_initial_metal) = 150°C
- For the water:
- Mass (m_water) = 100 g
- Initial temperature (T_initial_water) = 11°C
- Specific heat capacity (c_water) = 4.2 J g⁻¹ K⁻¹
- For the mixture:
- Final temperature (T_final) = 20°C
To find:
- Specific heat capacity of the metal (c_metal) = ?
Solution:
According to the principle of calorimetry, when no heat is lost to the surroundings, the heat lost by the hot body (the metal) is equal to the heat gained by the cold body (the water).
Heat lost by metal = Heat gained by water
The formula for heat energy (Q) transferred is given by:
Q = m × c × ΔT
where,
m = mass
c = specific heat capacity
ΔT = change in temperature
For the metal (hot body), the change in temperature is:
ΔT_metal = Initial temperature – Final temperature
=> ΔT_metal = 150°C – 20°C
=> ΔT_metal = 130°C
For the water (cold body), the change in temperature is:
ΔT_water = Final temperature – Initial temperature
=> ΔT_water = 20°C – 11°C
=> ΔT_water = 9°C
Now, we apply the principle of calorimetry:
Heat lost by metal = Heat gained by water
=> (m × c × ΔT)_metal = (m × c × ΔT)_water
=> 50 × c_metal × 130 = 100 × 4.2 × 9
=> 6500 × c_metal = 3780
=> c_metal = 3780 / 6500
=> c_metal = 0.58153… J g⁻¹ K⁻¹
Rounding to three significant figures, the specific heat capacity of the metal is 0.582 J g⁻¹ K⁻¹.
12. 45 g of water at 50°C in a beaker is cooled when 50 g of copper at 18°C is added to it. The contents are stirred till a final constant temperature is reached. Calculate the final temperature. The specific heat capacity of copper is 0.39 J g⁻¹ K⁻¹ and that of water is 4.2 J g⁻¹ K⁻¹. State the assumption used.
Answer:
To find:
The final temperature (T_f).
Solution:
According to the principle of calorimetry, when two bodies at different temperatures are in contact, the heat lost by the hotter body is equal to the heat gained by the colder body, assuming no heat is lost to the surroundings.
Heat lost by water = Heat gained by copper
The formula for heat energy (Q) transferred is:
Q = m × c × ΔT
where,
m = mass
c = specific heat capacity
ΔT = change in temperature
For water (hotter body):
Heat lost (Q_lost) = m_water × c_water × (Initial temperature – Final temperature)
=> Q_lost = 45 × 4.2 × (50 – T_f)
For copper (colder body):
Heat gained (Q_gained) = m_copper × c_copper × (Final temperature – Initial temperature)
=> Q_gained = 50 × 0.39 × (T_f – 18)
Now, setting Q_lost = Q_gained:
=> 45 × 4.2 × (50 – T_f) = 50 × 0.39 × (T_f – 18)
=> 189 × (50 – T_f) = 19.5 × (T_f – 18)
=> (189 × 50) – (189 × T_f) = (19.5 × T_f) – (19.5 × 18)
=> 9450 – 189T_f = 19.5T_f – 351
=> 9450 + 351 = 19.5T_f + 189T_f
=> 9801 = 208.5T_f
=> T_f = 9801 / 208.5
=> T_f = 47.007… °C
Rounding the answer to one decimal place, the final temperature is 47.0°C.
Assumption used:
The assumption is that no heat is lost to the surroundings. The system consisting of water and copper is considered to be perfectly isolated, meaning there is no heat transfer to or from the beaker, the stirrer, or the surrounding air.
13. 200 g of hot water at 80°C is added to 400 g of cold water at 10°C. Neglecting the heat taken by the container, calculate the final temperature of the mixture of water. Specific heat capacity of water = 4200 J kg⁻¹ K⁻¹.
Answer:
Given:
The data for the hot and cold water can be summarized in the following table. Note that the mass has been converted from grams (g) to kilograms (kg) to match the units of the specific heat capacity.
For Hot Water:
Mass (m_h) = 0.2 kg
Initial Temperature (T_h) = 80°C
For Cold Water:
Mass (m_c) = 0.4 kg
Initial Temperature (T_c) = 10°C
Specific heat capacity of water (c) = 4200 J kg⁻¹ K⁻¹
To find:
Final temperature of the mixture (T_f) = ?
Solution:
According to the principle of calorimetry, when no heat is lost to the surroundings, the heat lost by the hot body is equal to the heat gained by the cold body.
Heat lost by hot water = Heat gained by cold water
=> Q_lost = Q_gained
The formula for heat transferred (Q) is Q = m * c * ΔT, where m is mass, c is specific heat capacity, and ΔT is the change in temperature.
Let T_f be the final temperature of the mixture.
The equation can be written as:
m_h * c * (T_h – T_f) = m_c * c * (T_f – T_c)
Substituting the given values into the equation:
=> 0.2 * 4200 * (80 – T_f) = 0.4 * 4200 * (T_f – 10)
We can cancel out the specific heat capacity of water (4200) from both sides of the equation:
=> 0.2 * (80 – T_f) = 0.4 * (T_f – 10)
Now, we solve for T_f:
=> 16 – 0.2T_f = 0.4T_f – 4
=> 16 + 4 = 0.4T_f + 0.2T_f
=> 20 = 0.6T_f
=> T_f = 20 / 0.6
=> T_f = 200 / 6
=> T_f = 33.33 °C
Therefore, the final temperature of the mixture of water is 33.33°C.
14. The temperature of 600 g of cold water rises by 15°C when 300 g of hot water at 50°C is added to it. What was the initial temperature of the cold water?
Answer:
Given:
To find:
The initial temperature of the cold water (T_c) = ?
Formula:
According to the principle of calorimetry, when no heat is lost to the surroundings:
Heat lost by hot body = Heat gained by cold body
The formula for heat (Q) transferred is:
Q = m × c × ΔT
where,
m = mass
c = specific heat capacity
ΔT = change in temperature
Solution:
According to the principle of calorimetry:
Heat lost by hot water = Heat gained by cold water
=> m_h × c × ΔT_h = m_c × c × ΔT_c
Here, ‘c’ is the specific heat capacity of water, which is the same for both hot and cold water. So, it can be cancelled from both sides of the equation.
=> m_h × ΔT_h = m_c × ΔT_c
Let the final temperature of the mixture be T_f.
Let the initial temperature of the cold water be T_c.
For cold water, the temperature rises by 15°C.
=> ΔT_c = T_f – T_c = 15°C
=> T_f = T_c + 15 —(i)
For hot water, the temperature falls from 50°C to T_f.
=> ΔT_h = 50 – T_f —(ii)
Substituting the values in the equation m_h × ΔT_h = m_c × ΔT_c:
=> 300 × (50 – T_f) = 600 × 15
=> 300 × (50 – T_f) = 9000
=> 50 – T_f = 9000 / 300
=> 50 – T_f = 30
=> T_f = 50 – 30
=> T_f = 20°C
Now, we find the initial temperature of the cold water (T_c) using equation (i):
T_f = T_c + 15
=> 20 = T_c + 15
=> T_c = 20 – 15
=> T_c = 5°C
Therefore, the initial temperature of the cold water was 5°C.
15. 1.0 kg of water is contained in a 1.25 kW kettle. Calculate the time taken for the temperature of water to rise from 25°C to its boiling point 100°C. Specific heat capacity of water = 4.2 J g⁻¹ K⁻¹.
Answer:
Given:
Mass of water (m) = 1.0 kg
Power of kettle (P) = 1.25 kW
Initial temperature (T₁) = 25°C
Final temperature (T₂) = 100°C
Specific heat capacity of water (c) = 4.2 J g⁻¹ K⁻¹
To find:
Time taken (t) = ?
Solution:
First, we need to ensure all units are consistent for the calculation. We will convert mass from kg to g and power from kW to W.
Mass (m) = 1.0 kg = 1000 g
Power (P) = 1.25 kW = 1.25 × 1000 W = 1250 W (Note: 1 W = 1 J/s)
Next, we calculate the change in temperature (ΔT).
ΔT = Final temperature – Initial temperature
=> ΔT = 100°C – 25°C
=> ΔT = 75°C
(A change of 75°C is equal to a change of 75 K, so ΔT = 75 K)
Now, we calculate the heat energy (Q) required to raise the temperature of the water using the formula: Q = mcΔT.
Q = m × c × ΔT
=> Q = 1000 g × 4.2 J g⁻¹ K⁻¹ × 75 K
=> Q = 315000 J
Assuming the kettle is 100% efficient, the electrical energy (E) supplied by the kettle is equal to the heat energy (Q) absorbed by the water. The formula for electrical energy is E = P × t.
E = Q
=> P × t = 315000 J
Now, we can rearrange the formula to solve for time (t).
t = Q / P
=> t = 315000 J / 1250 J/s
=> t = 252 s
The time taken for the water to reach its boiling point is 252 seconds.
Exercise (B)
MCQ
1. Heat energy is ……………….. during melting and it is ……………….. during freezing at a constant temperature.
(a) rejected, absorbed
(b) rejected, rejected
(c) absorbed, absorbed
(d) absorbed, rejected
Answer: (d) absorbed, rejected
2. An increase in pressure results in ……………….. in melting point of ice:
(a) increase
(b) no change
(c) decrease
(d) none of the above
Answer: (c) decrease
3. At high altitudes, water boils at a temperature of :
(a) 100°C
(b) more than 100°C
(c) less than 100°C
(d) 150°C
Answer: (c) less than 100°C
4. If common salt is added to water, it boils at a temperature of:
(a) 100°C
(b) lower than 100°C
(c) higher than 100°C
(d) cannot say
Answer: (c) higher than 100°C
5. The specific latent heat of fusion of water is :
(a) 80 Cal g⁻¹
(b) 2260 J g⁻¹
(c) 80 J g⁻¹
(d) 336 J kg⁻¹
Answer: (a) 80 Cal g⁻¹
6. Heat energy supplied during the melting of a substance is utilised in :
(a) increasing the kinetic energy of molecules
(b) decreasing the potential energy of molecules
(c) increasing the potential energy of molecules
(d) decreasing the kinetic energy of molecules
Answer: (c) increasing the potential energy of molecules
Very Short Answer Type Questions
1. Write down the approximate range of temperature at which water boils in a pressure cooker.
Answer: The approximate range of temperature at which water boils in a pressure cooker is about 120°C to 125°C.
2. Complete the following sentences :
(a) When ice melts, its volume ……………….
(b) Decrease in pressure over ice ……………its melting point.
(c) Increase in pressure ……………… the boiling point of water.
(d) A pressure cooker is based on the principle that boiling point of water increases with the ……………...
(e) The boiling point of water is defined as ………………..
(f) Water can be made to boil at 115°C by ………………. pressure over its surface.
Answer: (a) When ice melts, its volume decreases.
(b) Decrease in pressure over ice increases its melting point.
(c) Increase in pressure increases the boiling point of water.
(d) A pressure cooker is based on the principle that boiling point of water increases with the increase in pressure.
(e) The boiling point of water is defined as the particular temperature at which vaporisation occurs.
(f) Water can be made to boil at 115°C by increasing pressure over its surface.
3. Write the approximate value of specific latent heat of ice.
Answer: The specific latent heat of fusion of ice is 336000 J kg⁻¹ (= 80 cal g⁻¹).
4. 1 g ice at 0°C melts to form 1 g water at 0°C. State whether the latent heat is absorbed or given out by ice.
Answer:
Given:
Initial State: 1 g of ice at 0°C.
Final State: 1 g of water at 0°C.
Process: Melting (change of state from solid to liquid at a constant temperature).
To find:
Whether the latent heat is absorbed or given out by the ice.
Solution:
The process described is the melting of ice into water. This is a change of state from solid to liquid.
For a substance to change from a solid state to a liquid state, its molecules need to gain energy. This energy is used to overcome the strong forces of attraction that hold the molecules in a fixed, rigid structure (the crystal lattice of ice).
This required energy is supplied in the form of heat from the surroundings. Since the ice must take in this heat energy to melt, it is an endothermic process.
The heat energy that is absorbed by a substance to change its state from solid to liquid at a constant temperature is called the latent heat of fusion.
Therefore, the latent heat is absorbed by the ice to change into water.
Short Answer Type Questions
1. A substance on heating, undergoes (i) a rise in its temperature, (ii) a change in its phase without change in its temperature. In each case, state the change in energy of molecules of the substance.
Answer: (i) When a substance on heating undergoes a rise in its temperature, the average kinetic energy of its molecules increases.
(ii) When a substance on heating undergoes a change in its phase without change in its temperature, the average potential energy of its molecules increases. For example, during melting, the heat energy supplied is utilised only in increasing the potential energy of the molecules.
2. How does the (a) average kinetic energy (b) average potential energy of molecules of a substance change during its change in phase at a constant temperature, on heating?
Answer: (a) During its change in phase at a constant temperature, on heating, the average kinetic energy of the molecules does not change.
(b) During its change in phase at a constant temperature, on heating, the average potential energy of the molecules on an average increases.
3. State the effect of presence of impurity on the melting point of ice. Give one use of it.
Answer: The melting point of a substance, like ice, decreases by the presence of impurities in it. This fact is utilised in making the freezing mixture by adding salt to ice. The freezing mixture is used in preparing kulfis.
4. State the effect of increase of pressure on the melting point of ice.
Answer: The melting point of ice decreases by the increase in pressure. For example, the melting point of ice decreases by 0.0072°C for every one atmosphere rise in pressure.
5. How is the boiling point of water affected when some salt is added to it?
Answer: The boiling point of a liquid, like water, increases by addition of impurities to it. If common salt is added to water, it boils at a temperature higher than 100°C.
6. What is the effect of increase of pressure on the boiling point of a liquid ?
Answer: The boiling point of a liquid increases with the increase in pressure and decreases with the decrease in pressure.
7. Water boils at 120°C in a pressure cooker. Explain the reason.
Answer: In a pressure cooker, steam is not allowed to escape out. The vapour pressure on water inside the pressure cooker becomes nearly 1.75 times the atmospheric pressure. This increases the boiling point of water so water boils in a cooker at about 120°C to 125°C due to increase in pressure.
8. It is difficult to cook vegetables on hills and mountains. Explain the reason.
Answer: At high altitudes, such as hills and mountains, the atmospheric pressure is low (less than one atmospheric pressure), therefore at these places, water boils at a temperature lower than 100°C and so it does not provide the required heat energy to its contents for cooking before it starts boiling. Thus, cooking at high altitudes, if done in the open, becomes difficult and takes a much longer time.
9. What do you understand by the term latent heat ?
Answer: The heat energy absorbed (or liberated) in change of phase is not externally manifested by any rise or fall in temperature, it is called the latent heat.
10. Define the term specific latent heat of fusion of ice. State its S.I. unit.
Answer: The specific latent heat of fusion of ice is the heat energy required to melt unit mass of ice at 0°C to water at 0°C without any change in temperature. The S.I. unit of specific latent heat is J kg⁻¹.
11. ‘The specific latent heat of fusion of ice is 336 J g⁻¹’. Explain the meaning of this statement.
Answer: ‘The specific latent heat of fusion of ice is 336 J g⁻¹’ means that 1 g of ice at 0°C absorbs 336 J of heat energy to convert into water at 0°C.
12. Which has more heat : 1 g of ice at 0°C or 1 g of water at 0°C? Give reason.
Answer: 1 g of water at 0°C has more heat. The reason is that 1 g of water at 0°C has 336 J (or 80 cal) heat energy more than 1 g of ice at 0°C. This additional energy is the latent heat of fusion absorbed by ice to change into water at 0°C.
13. (a) Which requires more heat : 1 g ice at 0°C or 1 g water at 0°C to raise its temperature to 10°C?
(b) Explain your answer in part (a).
Answer: (a) 1 g ice at 0°C requires more heat to raise its temperature to 10°C.
(b) 1 g of ice at 0°C first absorbs 336 J of heat energy (latent heat of fusion) to convert into 1 g of water at 0°C. Then this 1 g of water at 0°C absorbs heat energy to raise its temperature to 10°C. 1 g of water at 0°C only absorbs heat energy to raise its temperature to 10°C.
14. Ice cream appears colder to the mouth than water at 0°C. Give reason.
Answer: Ice cream appears colder to the mouth than water at 0°C because to melt, ice cream absorbs latent heat of fusion from the mouth. Thus, the mouth loses a greater amount of heat and feels colder. Water at 0°C does not absorb any such latent heat.
15. The soft drink bottles are cooled by (i) ice cubes at 0°C, and (ii) iced-water at 0°C. Which will cool the drink quickly? Give reason.
Answer: (i) Ice cubes at 0°C will cool the soft drink bottles more quickly. This is because 1 g of ice at 0°C takes 336 J of heat energy from the drink to melt into water at 0°C. Thus, the drink librates an additional 336 J of heat energy to 1 g ice at 0°C than to 1 g ice-cold water at 0°C. Therefore, cooling produced by 1 g ice at 0°C is much more than that by 1 g water at 0°C.
16. It is generally cold after a hail-storm than during and before the hail-storm. Give reason.
Answer: It is generally more cold after a hail-storm (when ice melts) than during or before the hail-storm. The reason is that after the hail-storm, ice starts absorbing the heat energy required for its melting from the surroundings, so the temperature of the surroundings falls further down and we feel more cold.
17. The temperature of the surrounding starts falling when ice in a frozen lake starts melting. Give reason.
Answer: When ice in a frozen lake starts melting, its surrounding becomes very cold. The reason is that quite a large amount of heat energy is required for melting the frozen lake which is absorbed from the surrounding atmosphere. As a result, the temperature of the surrounding falls and it becomes very cold.
18. Water in lakes and ponds do not freeze at once in cold countries. Give reason.
Answer: In cold countries water in lakes and ponds does not freeze all at once. The reason is that the specific latent heat of fusion of ice is sufficiently high (= 336 J g⁻¹). The water in lakes and ponds has to liberate a large quantity of heat to the surroundings (below 0°C) before freezing. The layer of ice formed over the water surface, being a poor conductor of heat, also prevents the loss of heat from the water of the lake, hence the water does not freeze all at once.
Long Answer Type Questions
1. (a) What do you understand by the change of phase of a substance ?
(b) Is there any change in temperature during the change of phase ?
(c) Does the substance absorb or liberate any heat energy during the change of phase ?
(d) What is the name given to the energy absorbed during a phase change ?
Answer: (a) The process of change from one state to another at a constant temperature is called the change of phase.
(b) No, there is no change in temperature during the change of phase. The change of phase occurs at a constant temperature.
(c) Yes, during the change of phase of a substance, a considerable amount of heat energy is absorbed or liberated. For example, heat energy is absorbed by a solid during melting and an equal amount of heat energy is liberated by the liquid during freezing.
(d) The energy absorbed during a phase change, since it is not externally manifested by any rise or fall in temperature, is called the latent heat.
2. A substance changes from its solid state to the liquid state when heat is supplied to it.
(a) Name the process.
(b) What name is given to heat absorbed by the substance.
(c) How does the average kinetic energy of molecules of the substance change ?
Answer: (a) The process is called melting.
(b) The heat absorbed by the substance is called the latent heat of fusion.
(c) The average kinetic energy of molecules of the substance does not change during melting because the temperature remains constant.
3. The diagram below shows the change of phases of a substance on a temperature-time graph on heating the substances at a constant rate.
(a) What do the parts AB, BC, CD and DE represent?
(b) What is the melting point of the substance ?
(c) What is the boiling point of the substance ?
Answer: (a) AB part represents the rise in temperature of the solid from temperature t₀ to t₁°C.
BC part represents melting at temperature t₁°C.
CD part represents the rise in temperature of the liquid from t₁°C to t₂°C.
DE part represents boiling at temperature t₂°C.
(b) The melting point of the substance is t₁°C.
(c) The boiling point of the substance is t₂°C.
4. The melting point of napthalene is 80°C and the room temperature is 25°. A sample of liquid napthalene at 90° is cooled down to room temperature. Draw a temperature-time graph to represent this cooling. On the graph mark the region which corresponds to the freezing process.
Answer: To represent the cooling of liquid naphthalene from 90°C to room temperature (25°C), a temperature-time graph would show temperature on the Y-axis and time on the X-axis.
Initially, the temperature would decrease from 90°C. At 80°C (the melting/freezing point), the temperature would remain constant for some time as the naphthalene freezes. This flat portion of the graph corresponds to the freezing process. After all the naphthalene has solidified, the temperature of the solid naphthalene would then decrease further from 80°C down to 25°C.
5. 1 kg of ice at 0°C is heated at a constant rate and its temperature is recorded after every 30 s till steam is formed at 100°C. Draw a temperature-time graph to represent the change of phases.
Answer: A temperature-time graph would be plotted with temperature on the Y-axis and time on the X-axis. The graph will show the following stages:
- A horizontal line at 0°C representing the melting of ice to water. During this phase, the temperature remains constant as ice absorbs latent heat of fusion.
- A rising line from 0°C to 100°C representing the heating of water. The temperature of water increases as it absorbs heat.
- A horizontal line at 100°C representing the boiling of water to steam. During this phase, the temperature remains constant as water absorbs latent heat of vaporization.
6. Explain the terms boiling and boiling point. How is the volume of water affected when it boils at 100°C?
Answer:
Boiling (or vaporisation) is the change from liquid to gas (or vapour) phase on absorption of heat at a constant temperature.
The boiling point of a liquid is the particular temperature at which vaporisation occurs.
When water boils at 100°C, its volume increases significantly. For example, 1 cm³ of water at 100°C becomes 1760 cm³ of steam at 100°C.
7. Explain the following :
(a) The surrounding become pleasantly warm when water in a lake starts freezing in cold countries.
(b) The heat supplied to a substance during its change of state, does not cause any rise in its temperature.
Answer:
(a) When water in a lake starts freezing in cold countries, it liberates a large quantity of heat (latent heat of fusion) to the surroundings without any change in its temperature. This liberated heat makes the surroundings pleasantly warm.
(b) The heat supplied to a substance during its change of state is used in increasing the potential energy of the molecules (i.e., for increasing the separation against the attractive forces between the molecules) and not in increasing their average kinetic energy. Since temperature is a measure of the average kinetic energy of molecules, it does not cause any rise in its temperature. This heat energy is called latent heat.
Numericals
1. 20 g of ice at 0°C absorbs 10,920 J of heat energy to melt and change to water at 50°C. Calculate the specific latent heat of fusion of ice. Specific heat capacity of water is 4200 J kg⁻¹ K⁻¹.
(i) Calculate the specific latent heat of fusion of ice.
Answer:
Given:
Mass of ice (m) = 20 g = 0.02 kg
Initial temperature (T_initial) = 0°C
Final temperature (T_final) = 50°C
Total heat energy absorbed (Q_total) = 10,920 J
Specific heat capacity of water (c) = 4200 J kg⁻¹ K⁻¹
To find:
Specific latent heat of fusion of ice (L_f) = ?
Solution:
The total heat absorbed is used in two stages:
- Heat required to melt the ice at 0°C into water at 0°C (Q1).
- Heat required to raise the temperature of the water from 0°C to 50°C (Q2).
Total Heat (Q_total) = Q1 + Q2
First, we calculate the heat required to raise the temperature of the water from 0°C to 50°C (Q2).
The formula for heat absorbed to change temperature is:
Q2 = m * c * ΔT
where,
m = 0.02 kg
c = 4200 J kg⁻¹ K⁻¹
ΔT = T_final – T_initial = 50°C – 0°C = 50°C (A change of 50°C is equal to a change of 50 K)
=> Q2 = 0.02 * 4200 * 50
=> Q2 = 84 * 50
=> Q2 = 4200 J
Now, we can find the heat used for melting the ice (Q1) using the total heat absorbed.
Q_total = Q1 + Q2
=> 10,920 = Q1 + 4200
=> Q1 = 10,920 – 4200
=> Q1 = 6720 J
The formula for heat required for a phase change (melting) is:
Q1 = m * L_f
We can now calculate the specific latent heat of fusion (L_f).
=> 6720 = 0.02 * L_f
=> L_f = 6720 / 0.02
=> L_f = 336,000 J/kg
Therefore, the specific latent heat of fusion of ice is 336,000 J/kg.
2. How much energy is released when 500 g of water at 80°C cools down to 0°C and then completely freezes ? [specific heat capacity of water = 4.2 J g⁻¹ K⁻¹, specific latent heat of fusion of ice = 336 J g⁻¹].
Answer:
Given:
Mass of water (m) = 500 g
Initial temperature (T_initial) = 80°C
Final temperature (T_final) = 0°C
Specific heat capacity of water (c) = 4.2 J g⁻¹ K⁻¹ (or 4.2 J g⁻¹ °C⁻¹)
Specific latent heat of fusion of ice (Lf) = 336 J g⁻¹
To find:
Total energy released (Q_total) = ?
Solution:
The total energy released is the sum of the energy released in two steps:
- Energy released when water cools from 80°C to 0°C (Q₁).
- Energy released when water at 0°C freezes into ice at 0°C (Q₂).
Step 1: Calculate the energy released during cooling (Q₁).
The formula for heat energy change due to a temperature change is:
Q₁ = m * c * ΔT
where ΔT is the change in temperature.
=> ΔT = Initial Temperature – Final Temperature
=> ΔT = 80°C – 0°C
=> ΔT = 80°C
Now,
=> Q₁ = 500 g × 4.2 J g⁻¹ °C⁻¹ × 80°C
=> Q₁ = 2100 × 80 J
=> Q₁ = 168000 J
Step 2: Calculate the energy released during freezing (Q₂).
The formula for heat energy change during a phase change (freezing) is:
Q₂ = m * Lf
Now,
=> Q₂ = 500 g × 336 J g⁻¹
=> Q₂ = 168000 J
Step 3: Calculate the total energy released (Q_total).
Q_total = Q₁ + Q₂
=> Q_total = 168000 J + 168000 J
=> Q_total = 336000 J
The total energy released is 336000 J or 336 kJ.
3. A molten metal of mass 150 g is kept at its melting point 800°C. When it is allowed to freeze at the same temperature, it gives out 75,000 J of heat energy.
(a) What is the specific latent heat of the metal ?
Answer:
Given:
Mass of the metal (m) = 150 g = 0.150 kg
Heat energy given out during freezing (Q) = 75,000 J
To find:
Specific latent heat of the metal (L) = ?
Solution:
The heat energy (Q) given out during freezing is related to the mass (m) and the specific latent heat of fusion (L) by the formula:
Q = m × L
To find the specific latent heat (L), we rearrange the formula:
L = Q / m
=> L = 75,000 J / 0.150 kg
=> L = 500,000 J kg⁻¹
Thus, the specific latent heat of the metal is 500,000 J kg⁻¹.
(b) If the specific heat capacity of metal is 200 J kg⁻¹ K⁻¹, how much additional heat energy will the metal give out in cooling to – 50 °C?
Answer:
Given:
Mass of the metal (m) = 0.150 kg
Specific heat capacity (c) = 200 J kg⁻¹ K⁻¹
Initial temperature (T_initial) = 800 °C (The metal is now solid at its freezing point)
Final temperature (T_final) = -50 °C
To find:
Additional heat energy given out (Q_additional) = ?
Solution:
First, we calculate the change in temperature (ΔT).
ΔT = Initial Temperature – Final Temperature
=> ΔT = 800 °C – (-50 °C)
=> ΔT = 800 + 50
=> ΔT = 850 °C
A change in temperature of 850 °C is equivalent to a change of 850 K.
The heat energy (Q) given out during cooling is calculated using the formula:
Q_additional = m × c × ΔT
=> Q_additional = 0.150 kg × 200 J kg⁻¹ K⁻¹ × 850 K
=> Q_additional = 30 × 850 J
=> Q_additional = 25,500 J
Thus, the additional heat energy the metal will give out in cooling to -50 °C is 25,500 J.
4. A solid metal of mass 150 g melts at its melting point of 800°C by providing heat at the rate of 100 W. The time taken for it to completely melt at the same temperature is 4 min. What is the specific latent heat of fusion of the metal ?
Answer:
Given:
Mass of the metal (m) = 150 g
Power of the heat source (P) = 100 W
Time taken to melt (t) = 4 min
Melting point = 800°C
To find:
Specific latent heat of fusion (L_f) = ?
Solution:
First, we will convert the given values into their standard SI units for consistency in calculation.
Mass (m) = 150 g = 150 / 1000 kg = 0.15 kg
Time (t) = 4 min = 4 × 60 s = 240 s
Power (P) = 100 W = 100 J/s
Now, we calculate the total heat energy (Q) supplied by the source to melt the metal. The relationship between power, energy (heat), and time is given by:
Q = P × t
=> Q = 100 J/s × 240 s
=> Q = 24000 J
This amount of heat is used entirely to change the state of the metal from solid to liquid at a constant temperature (the melting point). The formula for the heat required for a phase change (fusion) is:
Q = m × L_f
To find the specific latent heat of fusion (L_f), we can rearrange the formula:
L_f = Q / m
=> L_f = 24000 J / 0.15 kg
=> L_f = 160000 J/kg
The specific latent heat of fusion of the metal is 160,000 J/kg. This can also be expressed in scientific notation as 1.6 × 10^5 J/kg.
5. A refrigerator converts 100 g of water at 20°C to ice at -10°C in 73.5 min. Calculate the average rate of heat extraction in watt. The specific heat capacity of water is 4.2 J g⁻¹ K⁻¹, specific latent heat of ice is 336 J g⁻¹ and the specific heat capacity of ice is 2.1 J g⁻¹ K⁻¹.
Answer:
Given:
Mass of water (m) = 100 g
Initial temperature of water = 20°C
Final temperature of ice = -10°C
Time taken (t) = 73.5 min
Specific heat capacity of water (c_water) = 4.2 J g⁻¹ K⁻¹
Specific latent heat of ice (L_ice) = 336 J g⁻¹
Specific heat capacity of ice (c_ice) = 2.1 J g⁻¹ K⁻¹
To find:
The average rate of heat extraction (Power, P) in watt.
Solution:
The total heat extracted (Q_total) from the water to turn it into ice at -10°C occurs in three distinct stages:
- Heat extracted to cool the water from 20°C to 0°C (Q₁).
- Heat extracted to convert the water at 0°C to ice at 0°C (Q₂).
- Heat extracted to cool the ice from 0°C to -10°C (Q₃).
Step 1: Calculate the heat extracted to cool water from 20°C to 0°C (Q₁).
The formula for heat change is Q = m * c * ΔT.
=> Q₁ = m * c_water * (Change in temperature)
=> Q₁ = 100 g * 4.2 J g⁻¹ K⁻¹ * (20 – 0) K
=> Q₁ = 100 * 4.2 * 20
=> Q₁ = 8400 J
Step 2: Calculate the heat extracted during the phase change from water to ice at 0°C (Q₂).
The formula for latent heat is Q = m * L.
=> Q₂ = m * L_ice
=> Q₂ = 100 g * 336 J g⁻¹
=> Q₂ = 33600 J
Step 3: Calculate the heat extracted to cool ice from 0°C to -10°C (Q₃).
The formula for heat change is Q = m * c * ΔT.
=> Q₃ = m * c_ice * (Change in temperature)
=> Q₃ = 100 g * 2.1 J g⁻¹ K⁻¹ * (0 – (-10)) K
=> Q₃ = 100 * 2.1 * 10
=> Q₃ = 2100 J
Step 4: Calculate the total heat extracted (Q_total).
Q_total = Q₁ + Q₂ + Q₃
=> Q_total = 8400 J + 33600 J + 2100 J
=> Q_total = 44100 J
Step 5: Calculate the average rate of heat extraction (Power).
The rate of heat extraction (Power) is the total heat extracted divided by the time taken. The unit for power is watt (W), which is equivalent to joules per second (J/s).
Power (P) = Q_total / t
First, we need to convert the time from minutes to seconds.
=> t = 73.5 min
=> t = 73.5 * 60 s
=> t = 4410 s
Now, we can calculate the power.
=> P = 44100 J / 4410 s
=> P = 10 J/s
=> P = 10 W
Therefore, the average rate of heat extraction is 10 watts.
.6. In an experiment, 17 g of ice is used to bring down the temperature of 40 g of water at 34°C to its freezing temperature. The specific heat capacity of water is 4.2 J g⁻¹ K⁻¹. Calculate the specific latent heat of ice. State one important assumption made in the above calculation.
Answer:
Given:
To find:
Specific latent heat of ice (L_ice) = ?
Solution:
According to the principle of calorimetry, the heat lost by the hot body is equal to the heat gained by the cold body, assuming no heat is lost to the surroundings.
Heat lost by water = Heat gained by ice
Step 1: Calculate the heat lost by the water (Q_lost).
The water cools from 34°C to the final temperature of 0°C.
Change in temperature (ΔT) = Initial temperature – Final temperature
=> ΔT = 34°C – 0°C = 34°C
(A change of 34°C is equal to a change of 34 K)
Heat lost (Q_lost) = m_water × c_water × ΔT
=> Q_lost = 40 g × 4.2 J g⁻¹ K⁻¹ × 34 K
=> Q_lost = 5712 J
Step 2: Formulate the expression for heat gained by the ice (Q_gained).
The ice melts at 0°C to form water at 0°C. This is a phase change.
Heat gained (Q_gained) = m_ice × L_ice
=> Q_gained = 17 g × L_ice
Step 3: Equate heat lost and heat gained to find L_ice.
Q_lost = Q_gained
=> 5712 = 17 × L_ice
=> L_ice = 5712 / 17
=> L_ice = 336 J g⁻¹
Therefore, the specific latent heat of ice is 336 J g⁻¹.
Assumption:
One important assumption made in the above calculation is that no heat is lost to or gained from the surroundings. The system is considered to be perfectly isolated.
7. The temperature of 170 g of water at 50°C is lowered to 5°C by adding certain amount of ice to it. Find the mass of ice added. Given : Specific heat capacity of water = 4200 J kg⁻¹ °C⁻¹ and specific latent heat of ice = 336000 J kg⁻¹.
Answer:
Given:
Mass of water (m_w) = 170 g = 0.170 kg
Initial temperature of water (T_initial_w) = 50°C
Final temperature of the mixture (T_final) = 5°C
Specific heat capacity of water (c_w) = 4200 J kg⁻¹ °C⁻¹
Specific latent heat of ice (L_f) = 336000 J kg⁻¹
It is assumed that the ice is initially at 0°C.
To find:
Mass of ice added (m_ice) = ?
Solution:
According to the principle of calorimetry, when no heat is lost to the surroundings, the heat lost by the hotter body is equal to the heat gained by the colder body.
Heat lost = Heat gained
Step 1: Calculate the heat lost by the water.
The water cools from 50°C to the final temperature of 5°C.
Change in temperature of water (ΔT_w) = Initial temperature – Final temperature
=> ΔT_w = 50°C – 5°C
=> ΔT_w = 45°C
The formula for heat lost is: Q_lost = m_w × c_w × ΔT_w
=> Q_lost = 0.170 kg × 4200 J kg⁻¹ °C⁻¹ × 45°C
=> Q_lost = 32130 J
Step 2: Calculate the heat gained by the ice.
The ice gains heat in two stages:
(i) The ice at 0°C melts into water at 0°C by absorbing latent heat.
(ii) The resulting water from the melted ice at 0°C is heated to the final temperature of 5°C.
Let the mass of ice be m_ice.
(i) Heat gained by ice to melt (Q_melt) = m_ice × L_f
=> Q_melt = m_ice × 336000 J kg⁻¹
(ii) Heat gained by the melted ice (now water) to warm up from 0°C to 5°C (Q_warm).
Change in temperature of melted ice (ΔT_melted_ice) = 5°C – 0°C = 5°C
=> Q_warm = m_ice × c_w × ΔT_melted_ice
=> Q_warm = m_ice × 4200 J kg⁻¹ °C⁻¹ × 5°C
=> Q_warm = m_ice × 21000 J kg⁻¹
Total heat gained by ice (Q_gained) = Q_melt + Q_warm
=> Q_gained = (m_ice × 336000) + (m_ice × 21000)
=> Q_gained = m_ice × (336000 + 21000)
=> Q_gained = m_ice × 357000 J kg⁻¹
Step 3: Equate heat lost and heat gained to find the mass of ice.
Q_lost = Q_gained
=> 32130 J = m_ice × 357000 J kg⁻¹
=> m_ice = 32130 / 357000 kg
=> m_ice = 0.09 kg
To express the answer in grams:
=> m_ice = 0.09 × 1000 g
=> m_ice = 90 g
Thus, the mass of ice added is 90 g.
8. Find the result of mixing 10 g of ice at -10°C with 10 g of water at 10°C. Specific heat capacity of ice = 2.1 J g⁻¹ K⁻¹, specific latent heat of ice = 336 J g⁻¹ and specific heat capacity of water = 4.2 J g⁻¹ K⁻¹.
Answer:
Given:
The data for the two substances is summarized in the table below:
To find:
The final temperature and composition of the mixture.
Solution:
This problem is based on the principle of calorimetry, where the heat lost by the warmer substance is gained by the colder substance until they reach a state of thermal equilibrium. We will analyze the heat exchange in a step-by-step manner.
Step 1: Calculate the heat required to raise the temperature of the ice to its melting point (0°C).
Heat gained by ice to reach 0°C (Q₁) = m_ice × c_ice × ΔT_ice
=> Q₁ = 10 g × 2.1 J g⁻¹ K⁻¹ × (0 – (-10))°C
=> Q₁ = 10 × 2.1 × 10
=> Q₁ = 210 J.
Step 2: Calculate the maximum heat that the water can lose by cooling down to 0°C.
Heat lost by water to reach 0°C (Q₂) = m_water × c_water × ΔT_water
=> Q₂ = 10 g × 4.2 J g⁻¹ K⁻¹ × (10 – 0)°C
=> Q₂ = 10 × 4.2 × 10
=> Q₂ = 420 J.
Step 3: Compare the heat available from water (Q₂) with the heat required by ice (Q₁).
The heat that the water can release by cooling to 0°C (420 J) is greater than the heat required to warm the ice to 0°C (210 J).
=> Q₂ > Q₁
This indicates that the ice will reach 0°C, and the remaining heat from the water will be used to melt the ice.
Step 4: Calculate the heat required to melt all the ice at 0°C.
Heat required to melt all 10 g of ice (Q₃) = m_ice × L_f
=> Q₃ = 10 g × 336 J g⁻¹
=> Q₃ = 3360 J.
Step 5: Determine the final state of the mixture.
The total heat available from the water as it cools from 10°C to 0°C is 420 J. The heat required to warm the ice to 0°C is 210 J. The heat remaining to melt the ice is:
=> Heat available for melting = 420 J – 210 J = 210 J.
Since the heat available for melting (210 J) is less than the heat required to melt all the ice (3360 J), only a portion of the ice will melt. Therefore, the final temperature of the mixture will be 0°C.
Step 6: Calculate the mass of ice that melts.
The heat available for melting (210 J) will melt a certain mass of ice (m_melted).
=> Heat for melting = m_melted × L_f
=> 210 J = m_melted × 336 J g⁻¹
=> m_melted = 210 / 336
=> m_melted = 0.625 g.
Step 7: Determine the final composition of the mixture.
The final temperature of the mixture is 0°C. The final mixture consists of the remaining ice and the total amount of water.
Mass of remaining ice = Initial mass of ice – Mass of ice melted
=> Mass of remaining ice = 10 g – 0.625 g
=> Mass of remaining ice = 9.375 g.
Mass of total water = Initial mass of water + Mass of water from melted ice
=> Mass of total water = 10 g + 0.625 g
=> Mass of total water = 10.625 g.
Therefore, the result of mixing is a mixture containing 9.375 g of ice and 10.625 g of water at a final temperature of 0°C.
9. A piece of ice of mass 40 g is added to 200 g of water at 50°C. Calculate the final temperature of water when all the ice has melted. Specific heat capacity of water = 4200 J kg⁻¹ K⁻¹ and specific latent heat of fusion of ice = 336 × 10³ J kg⁻¹.
Answer:
Given:
To find:
The final temperature of the mixture (T_f) = ?
Solution:
According to the principle of calorimetry, when no heat is lost to the surroundings, the heat lost by the hot body is equal to the heat gained by the cold body.
Heat lost by hot water = Heat gained by ice
Let the final temperature of the mixture be T_f.
Heat lost by water = Mass of water × Specific heat capacity of water × Change in temperature
=> Q_lost = m_water × c_water × (50 – T_f)
Heat gained by ice occurs in two stages:
- Heat gained by ice to melt at 0°C (Latent heat of fusion).
- Heat gained by the melted ice (now water) to rise from 0°C to the final temperature T_f.
Heat gained to melt ice = Mass of ice × Specific latent heat of fusion of ice
=> Q_melt = m_ice × L_f
Heat gained by melted ice = Mass of ice × Specific heat capacity of water × Change in temperature
=> Q_warm = m_ice × c_water × (T_f – 0)
Therefore, the total heat gained is:
=> Q_gained = Q_melt + Q_warm
=> Q_gained = (m_ice × L_f) + (m_ice × c_water × T_f)
Now, equating the heat lost and heat gained:
Q_lost = Q_gained
=> m_water × c_water × (50 – T_f) = (m_ice × L_f) + (m_ice × c_water × T_f)
Substituting the given values:
=> 0.2 × 4200 × (50 – T_f) = (0.04 × 336 × 10³) + (0.04 × 4200 × T_f)
=> 840 × (50 – T_f) = 13440 + (168 × T_f)
=> 42000 – 840T_f = 13440 + 168T_f
=> 42000 – 13440 = 168T_f + 840T_f
=> 28560 = 1008T_f
=> T_f = 28560 / 1008
=> T_f = 28.33 °C
The final temperature of the water is 28.33°C.
10. Calculate the mass of ice needed to cool 150 g of water contained in a calorimeter of mass 50 g at 32°C such that the final temperature is 5°C. Specific heat capacity of calorimeter = 0.4 J g⁻¹ °C⁻¹, specific heat capacity of water = 4.2 J g⁻¹ °C⁻¹, latent heat capacity of ice = 330 J g⁻¹.
Answer:
Given:
The data for the substances involved in the heat exchange is summarized in the table below.
(as water)
Note: The initial temperature of ice is assumed to be 0°C as it is not specified.
To find:
Mass of ice (m_ice) = ?
Solution:
According to the principle of calorimetry, when no heat is lost to the surroundings, the heat lost by the hot bodies is equal to the heat gained by the cold bodies.
Heat Lost = Heat Gained
Step 1: Calculate the total heat lost by the hot water and the calorimeter.
The water and the calorimeter cool down from an initial temperature of 32°C to a final temperature of 5°C.
Change in temperature (ΔT) = Initial Temperature – Final Temperature
=> ΔT = 32°C – 5°C
=> ΔT = 27°C
Heat lost by water (Q_water) = m_water × c_water × ΔT
=> Q_water = 150 g × 4.2 J g⁻¹ °C⁻¹ × 27°C
=> Q_water = 17010 J
Heat lost by calorimeter (Q_cal) = m_calorimeter × c_calorimeter × ΔT
=> Q_cal = 50 g × 0.4 J g⁻¹ °C⁻¹ × 27°C
=> Q_cal = 540 J
Total Heat Lost (Q_lost) = Heat lost by water + Heat lost by calorimeter
=> Q_lost = 17010 J + 540 J
=> Q_lost = 17550 J
Step 2: Calculate the total heat gained by the ice.
Let the required mass of ice be ‘m_ice’.
The ice at 0°C first melts into water at 0°C, and then this water warms up to the final temperature of 5°C.
Heat gained to melt ice at 0°C (Q_melt) = m_ice × L_ice
=> Q_melt = m_ice × 330
Heat gained by the melted ice (water) to warm from 0°C to 5°C (Q_warm) = m_ice × c_water × (Final Temperature – 0°C)
=> Q_warm = m_ice × 4.2 × (5 – 0)
=> Q_warm = m_ice × 4.2 × 5
=> Q_warm = m_ice × 21
Total Heat Gained (Q_gained) = Heat gained to melt + Heat gained to warm
=> Q_gained = (m_ice × 330) + (m_ice × 21)
=> Q_gained = m_ice × (330 + 21)
=> Q_gained = m_ice × 351
Step 3: Equate the Total Heat Lost and Total Heat Gained to find the mass of ice.
Q_lost = Q_gained
=> 17550 = m_ice × 351
=> m_ice = 17550 / 351
=> m_ice = 50 g
Therefore, the mass of ice needed to cool the water to 5°C is 50 g.
11. 250 g of water at 30°C is contained in a copper vessel of mass 50 g. Calculate the mass of ice required to bring down the temperature of the vessel and its contents to 5°C. Given : specific latent heat of fusion of ice = 336 × 10³ J kg⁻¹, specific heat capacity of copper = 400 J kg⁻¹ K⁻¹, specific heat capacity of water = 4200 J kg⁻¹ K⁻¹.
Answer:
Given:
Mass of water (m_w) = 250 g = 0.250 kg
Mass of copper vessel (m_c) = 50 g = 0.050 kg
Initial temperature of water and vessel (T_initial) = 30°C
Final temperature of the mixture (T_final) = 5°C
Initial temperature of ice (T_ice) = 0°C (assumed)
Specific heat capacity of water (c_w) = 4200 J kg⁻¹ K⁻¹
Specific heat capacity of copper (c_c) = 400 J kg⁻¹ K⁻¹
Specific latent heat of fusion of ice (L_f) = 336 × 10³ J kg⁻¹ = 336000 J kg⁻¹
To find:
Mass of ice required (m_ice) = ?
Solution:
According to the principle of calorimetry, in an isolated system, the heat lost by the hotter bodies is equal to the heat gained by the colder bodies.
Heat Lost = Heat Gained
Step 1: Calculate the total heat lost by the copper vessel and the water.
The copper vessel and the water cool down from 30°C to 5°C.
Change in temperature (ΔT) = Initial Temperature – Final Temperature
=> ΔT = 30°C – 5°C = 25°C = 25 K
Heat lost by copper vessel (Q_c) = m_c × c_c × ΔT
=> Q_c = 0.050 kg × 400 J kg⁻¹ K⁻¹ × 25 K
=> Q_c = 500 J
Heat lost by water (Q_w) = m_w × c_w × ΔT
=> Q_w = 0.250 kg × 4200 J kg⁻¹ K⁻¹ × 25 K
=> Q_w = 26250 J
Total heat lost (Q_lost) = Heat lost by copper vessel + Heat lost by water
=> Q_lost = Q_c + Q_w
=> Q_lost = 500 J + 26250 J
=> Q_lost = 26750 J
Step 2: Calculate the total heat gained by the ice.
Let the mass of ice required be m_ice.
The ice at 0°C first melts into water at 0°C (latent heat of fusion), and then this melted water warms up to the final temperature of 5°C.
Heat gained by ice to melt at 0°C (Q_melt) = m_ice × L_f
=> Q_melt = m_ice × 336000 J/kg
Heat gained by the melted water to rise from 0°C to 5°C (Q_heat) = m_ice × c_w × (T_final – T_ice)
=> Q_heat = m_ice × 4200 J kg⁻¹ K⁻¹ × (5 – 0) K
=> Q_heat = m_ice × 4200 × 5
=> Q_heat = m_ice × 21000 J/kg
Total heat gained (Q_gained) = Q_melt + Q_heat
=> Q_gained = (m_ice × 336000) + (m_ice × 21000)
=> Q_gained = m_ice × (336000 + 21000)
=> Q_gained = m_ice × 357000 J
Step 3: Equate the heat lost and heat gained to find the mass of ice.
From the principle of calorimetry:
Q_lost = Q_gained
=> 26750 = m_ice × 357000
=> m_ice = 26750 / 357000
=> m_ice = 0.074929… kg
To express the mass in grams, we multiply by 1000:
=> m_ice = 0.074929… kg × 1000 g/kg
=> m_ice ≈ 74.93 g
Rounding off to two significant figures, the mass of ice required is 75 g.
The mass of ice required to bring down the temperature of the vessel and its contents to 5°C is 75 g.
12. 2 kg of ice melts when water at 100°C is poured in a hole drilled in a block of ice. What mass of water was used ? Given : specific heat capacity of water = 4200 J kg⁻¹ K⁻¹, specific latent heat of ice = 336 × 10³ J kg⁻¹.
Answer:
Given:
The following table lists the given quantities:
To find:
The following table shows the quantity to be found:
Solution:
This problem is based on the principle of calorimetry. According to this principle, when a hot body and a cold body are in thermal contact, the heat lost by the hot body is equal to the heat gained by the cold body, assuming no heat is lost to the surroundings.
Heat lost by hot water = Heat gained by ice
Step 1: Calculate the heat gained by the ice to melt.
The heat gained by the ice is used to change its state from solid at 0°C to liquid water at 0°C. This is calculated using the formula for latent heat of fusion.
Heat gained by ice (Q_gained) = Mass of ice × Specific latent heat of ice
=> Q_gained = m_ice × L_f
=> Q_gained = 2 kg × (336 × 10³ J kg⁻¹)
=> Q_gained = 672000 J
Step 2: Formulate the expression for heat lost by the water.
The hot water at 100°C cools down to the final temperature of the mixture, which is 0°C. The heat lost is calculated using the formula for specific heat capacity.
Change in temperature (ΔT) = Initial temperature – Final temperature
=> ΔT = 100°C – 0°C
=> ΔT = 100°C (or 100 K)
Heat lost by water (Q_lost) = Mass of water × Specific heat capacity of water × Change in temperature
=> Q_lost = m_w × c_w × ΔT
=> Q_lost = m_w × 4200 J kg⁻¹ K⁻¹ × 100 K
=> Q_lost = m_w × 420000 J
Step 3: Apply the principle of calorimetry to find the mass of water.
By equating the heat lost by the water to the heat gained by the ice, we can solve for the unknown mass of water (m_w).
Heat lost = Heat gained
=> Q_lost = Q_gained
=> m_w × 420000 = 672000
=> m_w = 672000 / 420000
=> m_w = 672 / 420
=> m_w = 1.6 kg
Thus, the mass of water used was 1.6 kg.
13. Calculate the total amount of heat energy required to convert 100 g of ice at -10°C completely into water at 100°C. Specific heat capacity of ice = 2.1 J g⁻¹ K⁻¹, specific heat capacity of water = 4.2 J g⁻¹ K⁻¹, specific latent heat of ice = 336 J g⁻¹.
Answer:
Given:
Mass of ice (m) = 100 g
Initial temperature of ice = -10°C
Final temperature of water = 100°C
Specific heat capacity of ice (c_ice) = 2.1 J g⁻¹ K⁻¹
Specific heat capacity of water (c_water) = 4.2 J g⁻¹ K⁻¹
Specific latent heat of ice (L_ice) = 336 J g⁻¹
To find:
Total heat energy required (Q_total) = ?
Solution:
The total heat energy required is the sum of the heat required for three separate processes:
- Heat energy to raise the temperature of ice from -10°C to 0°C (Q₁).
- Heat energy to melt the ice at 0°C into water at 0°C (Q₂).
- Heat energy to raise the temperature of water from 0°C to 100°C (Q₃).
Step 1: Calculate the heat energy to raise the temperature of ice from -10°C to 0°C (Q₁).
The formula for heat energy is Q = m × c × ΔT, where ΔT is the change in temperature.
ΔT = Final temperature – Initial temperature
=> ΔT = 0°C – (-10°C)
=> ΔT = 10°C
Now,
Q₁ = m × c_ice × ΔT
=> Q₁ = 100 g × 2.1 J g⁻¹ °C⁻¹ × 10°C
=> Q₁ = 2100 J
Step 2: Calculate the heat energy to melt the ice at 0°C (Q₂).
The formula for latent heat of fusion is Q = m × L.
Q₂ = m × L_ice
=> Q₂ = 100 g × 336 J g⁻¹
=> Q₂ = 33600 J
Step 3: Calculate the heat energy to raise the temperature of water from 0°C to 100°C (Q₃).
The formula for heat energy is Q = m × c × ΔT.
ΔT = Final temperature – Initial temperature
=> ΔT = 100°C – 0°C
=> ΔT = 100°C
Now,
Q₃ = m × c_water × ΔT
=> Q₃ = 100 g × 4.2 J g⁻¹ °C⁻¹ × 100°C
=> Q₃ = 42000 J
Step 4: Calculate the total heat energy (Q_total).
Q_total = Q₁ + Q₂ + Q₃
=> Q_total = 2100 J + 33600 J + 42000 J
=> Q_total = 77700 J
Therefore, the total amount of heat energy required is 77700 J.
14. The amount of heat energy required to convert 1 kg of ice at -10°C to water at 100°C is 7,77,000 J. Calculate the specific latent heat of ice. Specific heat capacity of ice = 2100 J kg⁻¹ K⁻¹, specific heat capacity of water = 4200 J kg⁻¹ K⁻¹.
Answer:
Given:
Total heat energy (Q_total) = 7,77,000 J
Mass of substance (m) = 1 kg
Initial temperature of ice (T_i) = -10°C
Final temperature of water (T_f) = 100°C
Specific heat capacity of ice (c_ice) = 2100 J kg⁻¹ K⁻¹
Specific heat capacity of water (c_water) = 4200 J kg⁻¹ K⁻¹
To find:
Specific latent heat of ice (L_ice) = ?
Solution:
The total heat energy supplied is used in three stages:
- Heat energy to raise the temperature of 1 kg of ice from -10°C to 0°C (Q1).
- Heat energy to convert 1 kg of ice at 0°C to water at 0°C (Q2). This is the latent heat of fusion.
- Heat energy to raise the temperature of 1 kg of water from 0°C to 100°C (Q3).
The total heat energy is the sum of the energy from these three stages:
Q_total = Q1 + Q2 + Q3
Step 1: Calculate Q1 (Heat to raise the temperature of ice)
Q1 = m × c_ice × ΔT_ice
=> Q1 = 1 × 2100 × (0 – (-10))
=> Q1 = 1 × 2100 × 10
=> Q1 = 21,000 J
Step 2: Calculate Q3 (Heat to raise the temperature of water)
Q3 = m × c_water × ΔT_water
=> Q3 = 1 × 4200 × (100 – 0)
=> Q3 = 1 × 4200 × 100
=> Q3 = 4,20,000 J
Step 3: Calculate the specific latent heat of ice (L_ice)
We know that Q2 is the heat required for melting, given by the formula Q2 = m × L_ice.
Now, we substitute the values of Q1 and Q3 into the total heat equation.
Q_total = Q1 + Q2 + Q3
=> Q_total = Q1 + (m × L_ice) + Q3
=> 7,77,000 = 21,000 + (1 × L_ice) + 4,20,000
=> 7,77,000 = 4,41,000 + L_ice
=> L_ice = 7,77,000 – 4,41,000
=> L_ice = 3,36,000 J kg⁻¹
The specific latent heat of ice is 3,36,000 J kg⁻¹.
15. 200 g of ice at 0°C converts into water at 0°C in 1 minute when heat is supplied to it at a constant rate. In how much time 200 g of water at 0°C will change to 20°C ? Take specific latent heat of ice = 336 J g⁻¹.
Answer:
Given:
For the melting of ice (Process 1):
Mass of ice (m) = 200 g
Initial temperature = 0°C
Final temperature = 0°C (as water)
Time taken (t₁) = 1 minute = 60 s
Specific latent heat of ice (L) = 336 J g⁻¹
For the heating of water (Process 2):
Mass of water (m) = 200 g
Initial temperature = 0°C
Final temperature = 20°C
Specific heat capacity of water (c) = 4.2 J g⁻¹ °C⁻¹
To find:
Time taken for 200 g of water at 0°C to change to 20°C (t₂).
Solution:
The problem involves two stages. First, we will calculate the rate at which heat is supplied to the ice. Second, we will use this rate to find the time required to heat the water.
Step 1: Calculate the rate of heat supply (P).
The heat required to melt the ice (Q₁) is given by the formula:
Q₁ = m × L
Substituting the given values:
=> Q₁ = 200 g × 336 J g⁻¹
=> Q₁ = 67200 J
This amount of heat is supplied in 1 minute (60 seconds). Since the heat is supplied at a constant rate (P), we have:
P = Heat supplied / Time taken
=> P = Q₁ / t₁
=> P = 67200 J / 60 s
=> P = 1120 J/s
So, the rate of heat supply is 1120 J/s.
Step 2: Calculate the time required to heat the water (t₂).
The heat required to raise the temperature of water from 0°C to 20°C (Q₂) is given by the formula:
Q₂ = m × c × ΔT
where ΔT is the change in temperature.
ΔT = Final temperature – Initial temperature
=> ΔT = 20°C – 0°C
=> ΔT = 20°C
Now, substituting the values into the formula for Q₂:
(We use the standard value for the specific heat capacity of water, c = 4.2 J g⁻¹ °C⁻¹)
=> Q₂ = 200 g × 4.2 J g⁻¹ °C⁻¹ × 20°C
=> Q₂ = 16800 J
The time taken (t₂) to supply this amount of heat at the constant rate (P) is:
t₂ = Heat required / Rate of heat supply
=> t₂ = Q₂ / P
=> t₂ = 16800 J / 1120 J/s
=> t₂ = 15 s
Therefore, it will take 15 seconds for 200 g of water at 0°C to change to 20°C.
16. During exercise, the body loses heat through evaporation of sweat. If a person loses 1 kg of sweat during exercise, how much energy does the body loose through evaporation ? How does the cooling effect of evaporation compare to heat loss due to specific heat capacity ? [Latent heat of vaporization = 2268 × 10³ J kg⁻¹, specific heat capacity of water = 4.2 × 10³ J kg⁻¹K⁻¹]
Answer: This problem has two parts. First, we will calculate the energy lost through the evaporation of 1 kg of sweat. Second, we will compare this to the heat loss associated with the specific heat capacity of water.
(i) Energy lost through evaporation
Given:
Mass of sweat (m) = 1 kg
Latent heat of vaporization of water (L_v) = 2268 × 10³ J kg⁻¹
To find:
Energy lost through evaporation (Q) = ?
Solution:
The energy lost when a substance changes its state from liquid to gas (evaporation) is calculated using the formula for latent heat of vaporization.
Q = m × L_v
=> Q = 1 kg × (2268 × 10³ J kg⁻¹)
=> Q = 2268 × 10³ J
=> Q = 2.268 × 10⁶ J
Thus, the body loses 2.268 × 10⁶ J of energy through the evaporation of 1 kg of sweat.
(ii) Comparison with heat loss due to specific heat capacity
To compare the cooling effects, we will calculate the amount of heat lost if the temperature of 1 kg of water drops by 1 K (or 1°C) and compare it to the heat lost by evaporating 1 kg of water.
Given:
Mass of water (m) = 1 kg
Specific heat capacity of water (c) = 4.2 × 10³ J kg⁻¹K⁻¹
Change in temperature (ΔT) = 1 K (assumed for a standard comparison)
To find:
The ratio of heat lost through evaporation to heat lost through a 1 K temperature drop.
Solution:
First, we calculate the heat lost (Q_c) due to the specific heat capacity for a 1 K temperature drop.
Q_c = m × c × ΔT
=> Q_c = 1 kg × (4.2 × 10³ J kg⁻¹K⁻¹) × 1 K
=> Q_c = 4.2 × 10³ J
Now, we compare the energy lost through evaporation (Q) with the energy lost due to the specific heat capacity (Q_c).
Comparison Ratio = Q / Q_c
=> Ratio = (2268 × 10³ J) / (4.2 × 10³ J)
=> Ratio = 2268 / 4.2
=> Ratio = 540
Conclusion:
The energy lost by evaporating 1 kg of sweat is 540 times greater than the energy lost when the temperature of the same amount of water drops by 1 K. This demonstrates that evaporation is a significantly more effective mechanism for cooling the body compared to simply reducing the temperature of the water on the skin.
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