A Level Physics 14.3 Specific Heat Capacity and Specific Latent Heat MCQs – 30 Practice Questions with Answers (Cambridge 9702)
This page covers Section 14.3 of the Cambridge International AS & A Level Physics (9702) syllabus, an A Level topic on thermal physics. It tests specific heat capacity, specific latent heat of fusion and vaporisation, heating curves, and mixed problems that combine temperature changes with changes of state.
Formula recap: Energy for a temperature change: E = m c Δθ. Energy for a change of state: E = m L. Electrical energy supplied: E = P t. See the full revision list on the A Level Physics Formula Sheet.
Specific Heat Capacity
Q1. The specific heat capacity of a substance is defined as the energy required to
- (A) raise the temperature of the substance by 1 K
- (B) raise the temperature of unit mass (1 kg) of the substance by 1 K
- (C) change the state of unit mass of the substance at constant temperature
- (D) raise the temperature of 1 mol of the substance by 1 K
Show Answer
Answer: B – Specific heat capacity c is the energy needed to raise the temperature of 1 kg of the substance by 1 K (equivalently, by 1 °C).
Q2. What is the unit of specific heat capacity?
- (A) J kg−1
- (B) J K−1
- (C) J kg−1 K
- (D) J kg−1 K−1
Show Answer
Answer: D – From c = E / (m Δθ), the unit is joules per kilogram per kelvin, J kg−1 K−1.
Q3. How much energy is needed to raise the temperature of 2.0 kg of water by 20 K? The specific heat capacity of water is 4200 J kg−1 K−1.
- (A) 84 kJ
- (B) 336 kJ
- (C) 168 kJ
- (D) 420 kJ
Show Answer
Answer: C – E = m c Δθ = 2.0 × 4200 × 20 = 168 000 J = 168 kJ.
Q4. A 100 W electric heater is used to heat 0.50 kg of a liquid for 120 s. The temperature of the liquid rises by 15 K. Assuming no energy is lost to the surroundings, the specific heat capacity of the liquid is
- (A) 1600 J kg−1 K−1
- (B) 800 J kg−1 K−1
- (C) 2400 J kg−1 K−1
- (D) 3200 J kg−1 K−1
Show Answer
Answer: A – Energy supplied E = P t = 100 × 120 = 12 000 J. Then c = E / (m Δθ) = 12 000 / (0.50 × 15) = 1600 J kg−1 K−1.
Q5. In the experiment of Q4, energy is in fact lost to the container and the surroundings, but the student still calculates c using c = P t / (m Δθ). The calculated value of c will be
- (A) smaller than the true value, because the liquid cools during heating
- (B) equal to the true value, because the losses cancel out
- (C) smaller than the true value, because not all of the energy reaches the liquid
- (D) larger than the true value, because the energy supplied P t is greater than the energy actually gained by the liquid
Show Answer
Answer: D – Only part of the electrical energy heats the liquid; the rest heats the container and is lost to the surroundings. Using the full P t overstates the energy absorbed by the liquid, so the calculated c is too large. Good insulation and a lid reduce this error.
Q6. A certain quantity of energy raises the temperature of 1.0 kg of a liquid by 24 K. The same quantity of energy is supplied to 3.0 kg of the same liquid. The temperature rise is
- (A) 24 K
- (B) 8 K
- (C) 12 K
- (D) 72 K
Show Answer
Answer: B – For fixed E and c, Δθ = E / (m c) is inversely proportional to the mass. Tripling the mass divides the rise by 3: 24 / 3 = 8 K.
Q7. An aluminium block of mass 0.40 kg is heated from 20 °C to 80 °C. The specific heat capacity of aluminium is 900 J kg−1 K−1. The energy supplied to the block is
- (A) 5.4 kJ
- (B) 10.8 kJ
- (C) 21.6 kJ
- (D) 54 kJ
Show Answer
Answer: C – Δθ = 60 K. E = m c Δθ = 0.40 × 900 × 60 = 21 600 J = 21.6 kJ.
Q8. Energy of 42 kJ is supplied to 1.0 kg of water. The specific heat capacity of water is 4200 J kg−1 K−1. The temperature rise of the water is
- (A) 10 K
- (B) 20 K
- (C) 5 K
- (D) 100 K
Show Answer
Answer: A – Δθ = E / (m c) = 42 000 / (1.0 × 4200) = 10 K.
Q9. Water is widely used as a coolant in car engines and in central heating systems mainly because
- (A) it has a low specific heat capacity, so it heats up very quickly
- (B) it is a good conductor of thermal energy
- (C) it expands a lot when heated
- (D) its large specific heat capacity means it can absorb or release a large amount of energy with only a small temperature change
Show Answer
Answer: D – With c = 4200 J kg−1 K−1, each kilogram of water carries a large amount of energy per kelvin, so it removes heat from an engine effectively and moderates coastal climates.
Q10. Two blocks X and Y of the same mass are given equal amounts of energy. The temperature of X rises by 5 K and the temperature of Y rises by 15 K. It follows that
- (A) X and Y have the same specific heat capacity
- (B) the specific heat capacity of X is three times that of Y
- (C) the specific heat capacity of Y is three times that of X
- (D) the specific heat capacity of X is one third that of Y
Show Answer
Answer: B – With E and m the same, c is inversely proportional to Δθ. X warms by one third as much, so cX = 3 cY.
Specific Latent Heat
Q11. The specific latent heat of a substance is defined as the energy required to
- (A) change the state of unit mass of the substance at constant temperature
- (B) raise the temperature of unit mass of the substance by 1 K
- (C) change the state of the substance at its boiling point only
- (D) raise the temperature of the substance from 0 °C to 100 °C
Show Answer
Answer: A – Specific latent heat L is the energy per unit mass needed to change state (for example solid to liquid) with no change in temperature.
Q12. What is the unit of specific latent heat?
- (A) J
- (B) J K−1
- (C) J kg−1
- (D) J kg−1 K−1
Show Answer
Answer: C – From L = E / m, the unit is joules per kilogram, J kg−1. There is no temperature change during the change of state, so no kelvin appears in the unit.
Q13. What energy is needed to melt 0.20 kg of ice at 0 °C? The specific latent heat of fusion of ice is 3.3 × 105 J kg−1.
- (A) 33 kJ
- (B) 66 kJ
- (C) 660 kJ
- (D) 6.6 kJ
Show Answer
Answer: B – E = m L = 0.20 × 3.3 × 105 = 6.6 × 104 J = 66 kJ.
Q14. What energy is needed to boil away 0.050 kg of water at 100 °C? The specific latent heat of vaporisation of water is 2.3 × 106 J kg−1.
- (A) 46 kJ
- (B) 230 kJ
- (C) 11.5 kJ
- (D) 115 kJ
Show Answer
Answer: D – E = m L = 0.050 × 2.3 × 106 = 1.15 × 105 J = 115 kJ.
Q15. For water, the specific latent heat of vaporisation (about 2.3 × 106 J kg−1) is much larger than the specific latent heat of fusion (about 3.3 × 105 J kg−1). This is because vaporisation requires
- (A) the molecules to be completely separated from each other against intermolecular forces, whereas melting only loosens the solid structure
- (B) a much higher temperature than melting
- (C) the molecules to move much faster in steam than in water
- (D) more mass to be changed than in melting
Show Answer
Answer: A – In melting, the molecules stay close together and only the regular solid arrangement is broken down. In vaporisation the molecules must be pulled completely apart against the attractive intermolecular forces, which takes much more energy.
Q16. While a solid is melting, energy is supplied at a steady rate but the temperature stays constant. The supplied energy is used to
- (A) increase the average kinetic energy of the molecules
- (B) make the molecules vibrate faster
- (C) increase the potential energy of the molecules by separating them against intermolecular forces
- (D) increase both the kinetic and potential energy of the molecules equally
Show Answer
Answer: C – Temperature depends on the average kinetic energy of the molecules, which does not change during melting. The energy increases the molecular potential energy as the molecules are moved apart against intermolecular attractions.
Q17. Energy of 6.6 × 105 J is supplied to ice at 0 °C. The specific latent heat of fusion of ice is 3.3 × 105 J kg−1. The mass of ice melted is
- (A) 0.50 kg
- (B) 1.0 kg
- (C) 3.0 kg
- (D) 2.0 kg
Show Answer
Answer: D – m = E / L = 6.6 × 105 / 3.3 × 105 = 2.0 kg.
Q18. A 250 W heater boils water at 100 °C. In 200 s, 0.020 kg of water is boiled away. The specific latent heat of vaporisation of water is
- (A) 2.5 × 105 J kg−1
- (B) 2.5 × 106 J kg−1
- (C) 1.0 × 106 J kg−1
- (D) 5.0 × 106 J kg−1
Show Answer
Answer: B – E = P t = 250 × 200 = 50 000 J. L = E / m = 50 000 / 0.020 = 2.5 × 106 J kg−1.
Q19. Steam at 100 °C condenses on a cold surface. How much energy is released when 0.10 kg of steam condenses to water at 100 °C? The specific latent heat of vaporisation of water is 2.3 × 106 J kg−1.
- (A) 230 kJ
- (B) 23 kJ
- (C) 42 kJ
- (D) 460 kJ
Show Answer
Answer: A – Condensation releases the same latent heat per kilogram that vaporisation absorbs. E = m L = 0.10 × 2.3 × 106 = 2.3 × 105 J = 230 kJ. This is why steam scalds are so severe.
Q20. The specific latent heat of fusion of ice is 3.3 × 105 J kg−1 and the specific latent heat of vaporisation of water is 2.3 × 106 J kg−1. The energy needed to vaporise 1.0 kg of water at 100 °C is, compared with the energy needed to melt 1.0 kg of ice at 0 °C, about
- (A) the same
- (B) twice as large
- (C) seven times as large
- (D) seventy times as large
Show Answer
Answer: C – Ratio = 2.3 × 106 / 3.3 × 105 ≈ 7. Vaporising a kilogram of water takes about seven times the energy needed to melt a kilogram of ice.
Heating Curves & Mixed Problems
Q21. A substance is heated at a constant rate and its temperature is plotted against time. On the heating curve, a horizontal (flat) section shows that the substance is
- (A) not absorbing any energy
- (B) changing state at constant temperature
- (C) being heated at a lower rate
- (D) losing energy as fast as it gains it
Show Answer
Answer: B – On a plateau the substance keeps absorbing energy, but the energy changes its state (increasing molecular potential energy), so the temperature remains constant until the change of state is complete.
Q22. On the same heating curve, the sloped sections show the temperature rising. For a fixed rate of heating, a steeper slope indicates
- (A) a larger specific latent heat
- (B) a larger mass of substance
- (C) a change of state taking place
- (D) a smaller thermal capacity (m c) of the sample
Show Answer
Answer: D – With constant power P, the rate of temperature rise is P / (m c). A steeper slope means a larger rate of rise, so m c must be smaller.
Q23. Two samples of the same mass are heated at the same constant rate. Sample P has a much longer melting plateau than sample Q. This shows that P has
- (A) a larger specific latent heat of fusion than Q
- (B) a larger specific heat capacity than Q
- (C) a higher melting point than Q
- (D) a smaller specific latent heat of fusion than Q
Show Answer
Answer: A – At the same heating power, the energy used in melting is P × (plateau time). A longer plateau means more energy per kilogram is needed, so the specific latent heat of fusion is larger.
Q24. Calculate the total energy needed to change 0.10 kg of ice at −10 °C into water at 20 °C. Specific heat capacity of ice = 2100 J kg−1 K−1; specific latent heat of fusion of ice = 3.3 × 105 J kg−1; specific heat capacity of water = 4200 J kg−1 K−1.
- (A) 10.5 kJ
- (B) 41.4 kJ
- (C) 43.5 kJ
- (D) 52.5 kJ
Show Answer
Answer: C – Warm the ice: 0.10 × 2100 × 10 = 2100 J. Melt the ice: 0.10 × 3.3 × 105 = 33 000 J. Warm the water: 0.10 × 4200 × 20 = 8400 J. Total = 2100 + 33 000 + 8400 = 43 500 J = 43.5 kJ.
Q25. Calculate the energy needed to change 0.20 kg of water at 20 °C completely into steam at 100 °C. Specific heat capacity of water = 4200 J kg−1 K−1; specific latent heat of vaporisation of water = 2.3 × 106 J kg−1.
- (A) 4.6 × 105 J
- (B) 5.3 × 105 J
- (C) 6.7 × 104 J
- (D) 5.9 × 105 J
Show Answer
Answer: B – Heat the water to 100 °C: 0.20 × 4200 × 80 = 67 200 J. Vaporise it: 0.20 × 2.3 × 106 = 460 000 J. Total = 67 200 + 460 000 = 527 200 J ≈ 5.3 × 105 J.
Q26. A liquid cools by evaporation because
- (A) the slowest molecules escape from the surface first
- (B) evaporation increases the average kinetic energy of the remaining molecules
- (C) the escaping molecules carry away no energy
- (D) the fastest molecules escape, so the average kinetic energy (and hence the temperature) of the remaining molecules falls
Show Answer
Answer: D – Only the most energetic molecules can escape from the surface. Their removal lowers the average kinetic energy of those left behind, which is why evaporation produces cooling (as with sweat on skin).
Q27. A metal block of mass 0.50 kg at 100 °C is placed in 0.30 kg of water at 20 °C. The specific heat capacity of the metal is 400 J kg−1 K−1 and that of water is 4200 J kg−1 K−1. Assuming no energy is lost to the surroundings, the final temperature is about
- (A) 31 °C
- (B) 40 °C
- (C) 60 °C
- (D) 25 °C
Show Answer
Answer: A – Energy lost by metal = energy gained by water: 0.50 × 400 × (100 − θ) = 0.30 × 4200 × (θ − 20). So 200(100 − θ) = 1260(θ − 20), giving 20 000 + 25 200 = 1460 θ, so θ ≈ 31 °C.
Q28. Water of mass 2.0 kg at 60 °C is mixed with 1.0 kg of water at 20 °C. With no energy losses, the final temperature of the mixture is
- (A) 30 °C
- (B) 40 °C
- (C) 47 °C
- (D) 50 °C
Show Answer
Answer: C – Energy lost by hot water = energy gained by cold water, and c cancels: 2.0(60 − θ) = 1.0(θ − 20). So 120 − 2θ = θ − 20, giving θ = 140 / 3 ≈ 47 °C. The result is nearer 60 °C because there is more hot water.
Q29. Ice at 0 °C is melted by a 200 W heater. It takes 250 s to melt 0.15 kg of ice completely. The specific latent heat of fusion of ice is
- (A) 3.3 × 104 J kg−1
- (B) 3.3 × 105 J kg−1
- (C) 1.3 × 106 J kg−1
- (D) 2.2 × 105 J kg−1
Show Answer
Answer: B – E = P t = 200 × 250 = 50 000 J. L = E / m = 50 000 / 0.15 ≈ 3.3 × 105 J kg−1.
Q30. A 2000 W electric kettle heats 1.0 kg of water from 20 °C to 100 °C. The specific heat capacity of water is 4200 J kg−1 K−1. Assuming no energy losses, the time taken is
- (A) 84 s
- (B) 120 s
- (C) 336 s
- (D) 168 s
Show Answer
Answer: D – E = m c Δθ = 1.0 × 4200 × 80 = 336 000 J. Time = E / P = 336 000 / 2000 = 168 s.
Keep practising: Work through more board-style questions on the Physics MCQs hub, and keep the key equations beside you with the A Level Physics Formula Sheet.
These are original practice questions written for revision purposes. They are not Cambridge past paper questions, and this page is not affiliated with or endorsed by Cambridge Assessment International Education.