CBSE Class 10 Physics Heat MCQs Set 31

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Multiple Choice Questions

Question 1: The water equivalent of a substance of mass 200 g is 400 g. What is the specific heat capacity of the substance (in cal g-1 °C-1)?
(a) 1
(b) 0.5
(c) 2
(d) 8
Show Answer & Explanation

Answer: (c) 2

Explanation:
1. Water equivalent W = m × s, so 400 = 200 × s.
2. s = 2 cal g-1 °C-1.

Teacher's Note:
1. Water equivalent has the unit of gram.

Question 2: A liquid of mass m at 20 °C is mixed with a similar liquid of mass m at 30 °C. What is the temperature of the mixture?
(a) 30 °C
(b) 20 °C
(c) 50 °C
(d) 25 °C
Show Answer & Explanation

Answer: (d) 25 °C

Explanation:
1. Heat lost = heat gained: ms(30 - T) = ms(T - 20).
2. T = 25 °C.

Teacher's Note:
1. Equal masses of the same liquid give the mean temperature.

Question 3: Which of the following statements defines heat?
(a) the internal energy of a substance
(b) the part of internal energy that is transferred
(c) the total kinetic energy of the molecules of a substance
(d) the total potential energy present in a body
Show Answer & Explanation

Answer: (b) the part of internal energy that is transferred

Explanation:
1. Heat is the energy that flows from a body at higher temperature to one at lower temperature.
2. So it is the part of the internal energy that is transferred.

Teacher's Note:
1. Heat is energy in transit.

Question 4: A substance of mass 400 g at 30 °C and specific heat capacity 1 cal g-1 °C-1 is dropped into a calorimeter containing 50 g of a liquid at 10 °C (specific heat capacity 2 cal g-1 °C-1). The final temperature is 20 °C. What is the water equivalent of the calorimeter?
(a) 500 g
(b) 100 g
(c) 300 g
(d) 200 g
Show Answer & Explanation

Answer: (c) 300 g

Explanation:
1. Heat lost by the substance = 400 × 1 × (30 - 20) = 4000 cal.
2. Heat gained by the liquid = 50 × 2 × 10 = 1000 cal, and by the calorimeter = W × 10.
3. 4000 = 1000 + 10W, so W = 300 g.

Teacher's Note:
1. Principle of calorimetry: heat lost = heat gained.

Question 5: A metal plate with a hole in it is heated. What happens to the size of the hole?
(a) increases
(b) decreases
(c) remains the same
(d) increases or decreases depending on the nature of the material
Show Answer & Explanation

Answer: (a) increases

Explanation:
1. On heating, every linear dimension expands, including the hole, as if it were made of the same metal.
2. So the hole becomes larger.

Teacher's Note:
1. A hole expands as if it were filled with the metal.

Question 6: Three metal rods of initial lengths ℓ1 > ℓ2 > ℓ3, made of different materials, are heated to the same temperature. What is the relation between the final lengths?
(a) ℓ1' > ℓ2' > ℓ3'
(b) ℓ1' < ℓ2' < ℓ3'
(c) ℓ1' = ℓ2' = ℓ3'
(d) The final lengths depend on the nature of the material.
Show Answer & Explanation

Answer: (d) The final lengths depend on the nature of the material.

Explanation:
1. The expansion ℓα(Δt) depends on the coefficient of linear expansion α of each material.
2. Without α no order can be fixed, so the final lengths depend on the materials.

Teacher's Note:
1. Different materials expand by different amounts.

Question 7: The time period of a pendulum clock is 1 s at 0 °C. If the temperature of the brass pendulum (α = 2 × 10-5 °C-1) is raised to 50 °C, what is the new time period?
(a) 1 s
(b) 1.0005 s
(c) 1.0001 s
(d) 1.02 s
Show Answer & Explanation

Answer: (b) 1.0005 s

Explanation:
1. ℓ' = ℓ(1 + αΔt) = ℓ(1 + 2 × 10-5 × 50) = 1.001ℓ.
2. T ∝ √ℓ, so T' = 1 × \( \sqrt{1.001} \) ≈ 1.0005 s.

Teacher's Note:
1. T ∝ √ℓ, so the fractional change in T is half that in ℓ.

Question 8: Water falls from a height of 300 m. If 80% of the mechanical energy is used to raise the temperature of the water, what is the rise in temperature (in °C)? (g = 10 m s-2, specific heat capacity of water = 4200 J kg-1 °C-1)
(a) 0.34
(b) 0.42
(c) 0.82
(d) 0.57
Show Answer & Explanation

Answer: (d) 0.57

Explanation:
1. 0.8 mgh = mcΔT, so ΔT = 0.8 × 10 × 300/4200.
2. ΔT = 2400/4200 = 0.57 °C.

Teacher's Note:
1. Mass cancels from both sides.

Question 9: 5 g of ice at 0 °C is mixed with 5 g of water at 20 °C. What is the final temperature (in °C)? (latent heat of fusion of ice = 80 cal g-1)
(a) 0
(b) 60
(c) 15
(d) 10
Show Answer & Explanation

Answer: (a) 0

Explanation:
1. Heat released by water cooling to 0 °C = 5 × 1 × 20 = 100 cal.
2. Melting all the ice needs 5 × 80 = 400 cal, which is more than available.
3. So only part of the ice melts and the mixture stays at 0 °C.

Teacher's Note:
1. Compare the available heat with the heat needed for melting.

Question 10: 5 kg of water at 70 °C is mixed with some water at 20 °C so that the resultant temperature is 40 °C. How much water (in kg) at 20 °C is added?
(a) 7.5
(b) 5
(c) 10
(d) 2.5
Show Answer & Explanation

Answer: (a) 7.5

Explanation:
1. Heat lost = 5 × 1 × (70 - 40) = 150 units.
2. Heat gained = x × 1 × (40 - 20) = 20x, so x = 7.5 kg.

Teacher's Note:
1. Use heat lost = heat gained.

Question 11: Match Column A with Column B.
(A) Specific heat capacity (B) Latent heat of fusion (C) Water equivalent
(P) gram (Q) cal g-1 °C-1 (R) cal g-1
(a) A → R; B → P; C → Q
(b) A → Q; B → R; C → P
(c) A → R; B → Q; C → P
(d) A → Q; B → P; C → R
Show Answer & Explanation

Answer: (b) A → Q; B → R; C → P

Explanation:
1. Specific heat capacity is measured in cal g-1 °C-1 (Q).
2. Latent heat is heat per unit mass, cal g-1 (R).
3. Water equivalent is a mass, so its unit is gram (P).

Teacher's Note:
1. Check the units to match.

Question 12: Arrange the steps to determine the specific heat capacity of a substance.
(A) Note the initial temperature of the water.
(B) Use the law of mixtures.
(C) Take a known mass of water in a calorimeter.
(D) Determine the specific heat capacity of the material.
(E) Drop the material taken from hot water into the calorimeter.
(F) Find the temperature of the mixture.
(a) C E A B F D
(b) C A E B F D
(c) D A E F B C
(d) C A E F B D
Show Answer & Explanation

Answer: (d) C A E F B D

Explanation:
1. Take the water (C) and note its temperature (A).
2. Drop in the hot material (E), find the final temperature (F).
3. Apply the law of mixtures (B) and determine the specific heat capacity (D).

Teacher's Note:
1. Do the measurements before the calculation.

Question 13: A body X is not in thermal equilibrium with A, and A is not in thermal equilibrium with B. Which statement holds?
(a) X and B cannot be in thermal equilibrium.
(b) X and B must be in thermal equilibrium.
(c) X and B may be in thermal equilibrium.
(d) X, A and B must have different temperatures.
Show Answer & Explanation

Answer: (c) X and B may be in thermal equilibrium.

Explanation:
1. X and B both differ in temperature from A, but they could both be hotter (or both colder) than A and so have the same temperature.
2. So X and B may or may not be in thermal equilibrium.

Teacher's Note:
1. The zeroth law only says that bodies in equilibrium with a third are in equilibrium with each other.

Question 14: An object that is completely immersed is placed in water at 20 °C and then in water at 40 °C. If the upthrust is F1 at 20 °C and F2 at 40 °C, what is the relation between them?
(a) F1 = F2
(b) F1 > F2
(c) F1 < F2
(d) F1 = 2F2
Show Answer & Explanation

Answer: (b) F1 > F2

Explanation:
1. Upthrust = V ρw g, where V is the volume of the object.
2. As water is heated its density falls much more than the object's volume rises, so the upthrust decreases and F1 > F2.

Teacher's Note:
1. Liquids expand more than solids on heating.

Question 15: The temperature of the atmosphere is 20 °C and the water vapour present is 12.78 g m-3. Using the table (saturation at 10 °C: 9.38; 15 °C: 12.78; 20 °C: 17.18; 25 °C: 22.88 g m-3), find the dew point.
(a) 20 °C
(b) 15 °C
(c) 10 °C
(d) 25 °C
Show Answer & Explanation

Answer: (b) 15 °C

Explanation:
1. The dew point is the temperature at which the existing vapour saturates the air.
2. The saturation value 12.78 g m-3 corresponds to 15 °C.

Teacher's Note:
1. Dew point: the temperature at which the air becomes saturated.

Question 16: The water equivalent of a substance is 40 g. How much heat (in cal) is needed to raise its temperature by 20 °C?
(a) 400
(b) 200
(c) 800
(d) 1600
Show Answer & Explanation

Answer: (c) 800

Explanation:
1. Heat = water equivalent × ΔT × 1 cal g-1 °C-1 = 40 × 20 = 800 cal.

Teacher's Note:
1. Water equivalent converts the body into an equivalent mass of water.

Question 17: A liquid at 50 °C is mixed with a similar liquid of the same mass and the resultant temperature is 30 °C. What is the initial temperature of the second liquid?
(a) 40 °C
(b) 30 °C
(c) 20 °C
(d) 10 °C
Show Answer & Explanation

Answer: (d) 10 °C

Explanation:
1. For equal masses of the same liquid, the final temperature is the mean: (50 + T)/2 = 30.
2. T = 10 °C.

Teacher's Note:
1. The final temperature is the average.

Question 18: Which of the following defines temperature?
(a) total kinetic energy of the molecules
(b) average kinetic energy of the molecules
(c) total internal energy of the molecules
(d) total potential energy of the molecules
Show Answer & Explanation

Answer: (b) average kinetic energy of the molecules

Explanation:
1. Temperature is a measure of the average kinetic energy of the molecules.

Teacher's Note:
1. Temperature does not depend on how many molecules there are.

Question 19: A container whose water equivalent is 200 g holds 300 g of water at 30 °C. How much heat (in cal) is required to raise the temperature of the contents to 70 °C?
(a) 10,000
(b) 20,000
(c) 40,000
(d) 80,000
Show Answer & Explanation

Answer: (b) 20,000

Explanation:
1. Total water equivalent = 200 + 300 = 500 g.
2. Heat = 500 × (70 - 30) = 20,000 cal.

Teacher's Note:
1. Add the water equivalent of the container to the mass of water.

Question 20: A hollow metal sphere is heated. What happens to its internal volume and external volume respectively?
(a) increases, increases
(b) decreases, decreases
(c) increases, decreases
(d) decreases, increases
Show Answer & Explanation

Answer: (a) increases, increases

Explanation:
1. The cavity expands as though it were filled with the same metal.
2. So both the internal and the external volumes increase.

Teacher's Note:
1. Cavities expand as if they were solid.

Question 21: Equal volumes of three different liquids A, B and C are heated through the same temperature change. If ΔVA, ΔVB and ΔVC are the changes in volume, which is true?
(a) ΔVA = ΔVB = ΔVC
(b) ΔVA > ΔVB > ΔVC
(c) ΔVA < ΔVB < ΔVC
(d) the change in volume depends on the nature of the liquid used
Show Answer & Explanation

Answer: (d) the change in volume depends on the nature of the liquid used

Explanation:
1. ΔV = Vγ Δt, and the coefficient of volume expansion γ is different for different liquids.
2. So the change depends on the liquid.

Teacher's Note:
1. Each substance has its own expansion coefficient.

Question 22: A pendulum has a time period of 2 s at 10 °C. What is its time period at 110 °C? (α = 2 × 10-5 °C-1)
(a) 2 s
(b) 2.005 s
(c) 1.008 s
(d) 2.002 s
Show Answer & Explanation

Answer: (d) 2.002 s

Explanation:
1. ℓ' = ℓ(1 + 2 × 10-5 × 100) = 1.002ℓ.
2. T' = 2 × \( \sqrt{1.002} \) ≈ 2 × 1.001 = 2.002 s.

Teacher's Note:
1. T ∝ √ℓ.

Question 23: Water at 20 °C falls from a height of 500 m, and 90% of the potential energy is used to raise its temperature. What is the final temperature (in °C)? (g = 10 m s-2, c = 4200 J kg-1 °C-1)
(a) 21.07
(b) 22.08
(c) 23.01
(d) 21.8
Show Answer & Explanation

Answer: (a) 21.07

Explanation:
1. 0.9 × 10 × 500 = 4500 J per kg raises the temperature by 4500/4200 = 1.07 °C.
2. Final temperature = 20 + 1.07 = 21.07 °C.

Teacher's Note:
1. Work per kilogram is gh.

Question 24: 10 g of ice at -20 °C is mixed with 10 g of water at 10 °C. What is the final temperature (in °C)? (specific heat capacity of ice = 0.5 cal g-1 °C-1, latent heat of fusion = 80 cal g-1)
(a) 5
(b) 10
(c) 0
(d) 20
Show Answer & Explanation

Answer: (c) 0

Explanation:
1. Heat released by the water cooling from 10 °C to 0 °C = 10 × 10 = 100 cal.
2. Heat needed to warm the ice from -20 °C to 0 °C = 10 × 0.5 × 20 = 100 cal.
3. These are exactly equal, so the mixture ends at 0 °C with no melting.

Teacher's Note:
1. Check whether the heat available just warms the ice to 0 °C.

Question 25: How many metal balls, each of mass 50 g and specific heat capacity 0.1 cal g-1 °C-1 at 90 °C, must be added to 300 g of water at 25 °C so that the final temperature is 60 °C?
(a) 50
(b) 70
(c) 90
(d) 120
Show Answer & Explanation

Answer: (b) 70

Explanation:
1. Each ball gives 50 × 0.1 × (90 - 60) = 150 cal.
2. The water takes 300 × 1 × (60 - 25) = 10,500 cal.
3. Number of balls = 10,500/150 = 70.

Teacher's Note:
1. Heat lost = heat gained.

Question 26: Match Column A with Column B.
(A) Heat (B) Temperature (C) Specific heat capacity
(P) average kinetic energy (Q) internal energy flowing (R) depends on the nature of the material
(a) A → Q; B → P; C → R
(b) A → R; B → Q; C → P
(c) A → P; B → R; C → Q
(d) A → Q; B → R; C → P
Show Answer & Explanation

Answer: (a) A → Q; B → P; C → R

Explanation:
1. Heat is internal energy in transit (Q).
2. Temperature measures the average kinetic energy (P).
3. Specific heat capacity depends on the material (R).

Teacher's Note:
1. Heat and temperature are different quantities.

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