Specific heat capacity (RP1) – GCSE Physics exam questions with mark scheme

1Grade 4–53 marks

A pan contains 0.50 kg of water at 20 °C. The water is heated to 80 °C.

specific heat capacity of water = 4200 J/kg °C

Calculate the energy transferred to the water.

Use the equation: change in thermal energy $=$ mass $\times$ specific heat capacity $\times$ temperature change [3]

temperature change $= 60$ (°C)(1)
$\Delta E = 0.50 \times 4200 \times 60$(1)
126 000 J allow 1.26 × 10⁵ J; allow 126 kJ(1)
Answer: 126 000 J
Examiner tip: Use the temperature change (60 °C), not the final temperature (80 °C).

2Grade 6–7Higher5 marks

A student heats a 1.2 kg metal block with an electric heater. The heater transfers 12 000 J of energy. The temperature of the block rises from 21.0 °C to 46.0 °C.

(a) Calculate the specific heat capacity of the metal.

Give your answer in J/kg °C. [3]

(b) The student's value is higher than the true value for the metal.

Explain why. [2]

(a) temperature change $= 25.0$ (°C)(1)
(a) $12\,000 = 1.2 \times c \times 25.0$ or $c = \dfrac{12\,000}{1.2 \times 25.0}$(1)
(a) 400 (J/kg °C) a correct answer with no working scores 3 marks(1)
(b) some energy (from the heater) is transferred to the surroundings / not all the energy goes into the block allow some energy heats the heater or thermometer; ignore 'human error'(1)
(b) so the temperature rise is smaller than it would be if all 12 000 J went into the block, (so the calculated value of $c$ is too large) allow so the energy used in the calculation (12 000 J) is greater than the energy that actually heated the block; allow (for both marks) the final temperature was read before the block reached its maximum temperature / poor thermal contact between thermometer and block, so the measured temperature rise is too small(1)
Answer: (a) 400 J/kg °C (b) energy lost to surroundings, so temperature rise is too small
Examiner tip: In (b) you must link the energy loss to a smaller temperature rise — just saying 'heat was lost' scores only 1.

3Grade 6–7Higher6 marks

A student wants to determine the specific heat capacity of an aluminium block.

She has an aluminium block with two holes drilled in it, a 12 V immersion heater, a power supply, a thermometer, a balance, a joulemeter (or an ammeter, voltmeter and stopwatch) and some insulation.

Describe a method she could use. Include how she would use her measurements to determine the specific heat capacity. [6]

measure the mass of the block (using a balance)(1)
place the heater in one hole and the thermometer in the other (add a drop of water/oil to the thermometer hole for good thermal contact)(1)
wrap the block in insulation (to reduce energy transfer to the surroundings)(1)
record the starting temperature(1)
switch on the heater and record the energy transferred with the joulemeter, or record the current and pd and use $E = VIt$ / $P = VI$, $E = Pt$(1)
record the temperature at regular intervals (e.g. every minute) for about 10 minutes(1)
calculate the temperature change(1)
use $c = \dfrac{\Delta E}{m \Delta\theta}$; or plot temperature (y) against energy transferred (x) and use $c = \dfrac{1}{m \times \text{gradient}}$(1)

Level	Marks	What it looks like
Level 3	5–6	A coherent, logically ordered method that could be followed to get valid results. Includes how energy and temperature change are measured and how they are used with the mass to calculate the specific heat capacity.
Level 2	3–4	The main steps of a method are given, measuring energy and temperature, but some detail is missing or the order is not fully logical, or how to calculate the specific heat capacity is unclear.
Level 1	1–2	Some relevant steps are given, but they are disjointed and would not lead to a valid result.
No marks	0	No relevant content.

Answer: Measure mass; insulate block; heater and thermometer in holes; record start temperature; measure energy transferred (joulemeter or $VIt$) and temperature rise; $c = \Delta E / (m \Delta\theta)$.
Examiner tip: Top-band answers end with how the measurements are combined — many good methods stop at 'record the temperature' and drop into Level 2.

4Grade 1–3Foundation4 marks

(a) Which statement defines the specific heat capacity of a substance?

Tick (✓) one box. [1]

☐ The energy needed to change 1 kg of the substance from a solid to a liquid.
☐ The energy needed to raise the temperature of 1 kg of the substance by 1 °C.
☐ The energy needed to raise the temperature of the substance by 1 °C.
☐ The total energy stored in 1 kg of the substance.

(b) Give the unit of specific heat capacity. [1]

Substance	Specific heat capacity in J/kg °C
glass	840
lead	130
olive oil	1970
water	4200

Equal masses of each substance are given the same amount of energy.

Which substance will have the largest increase in temperature? Give a reason for your answer. [2]

(a) The energy needed to raise the temperature of 1 kg of the substance by 1 °C (only this box ticked; more than one box ticked = 0)(1)
(b) J/kg °C (allow J/kg K or J kg⁻¹ °C⁻¹; do not accept J or J/kg alone)(1)
(c) lead(1)
(c) (because) it has the lowest specific heat capacity (allow it needs the least energy to raise the temperature of 1 kg by 1 °C; reason only scores if lead chosen)(1)
Answer: (a) 2nd box; (b) J/kg °C; (c) lead — lowest specific heat capacity
Examiner tip: In (c) say "lowest specific heat capacity" — a vague answer like "it is a metal" or "it is denser" does not score.

5Grade 4–54 marks

A plumber uses a blowtorch to heat a copper pipe joint.

The mass of the joint is 0.25 kg. The joint starts at a temperature of 18 °C.

The blowtorch transfers 4620 J of energy to the thermal energy store of the joint.

specific heat capacity of copper = 385 J/kg °C

Calculate the final temperature of the joint.

Use the Physics Equations Sheet. [4]

4620 = 0.25 × 385 × Δθ(1)
Δθ = 4620 / (0.25 × 385) (rearrangement, may be implied by correct answer)(1)
Δθ = 48 (°C)(1)
final temperature = 66 °C (allow ecf: their Δθ + 18 correctly evaluated; an answer of 48 °C scores 3 marks; 66 °C with no working scores 4 marks)(1)
Answer: 66 °C
Examiner tip: The equation gives the temperature change — many students stop at 48 °C and forget to add the starting temperature.

6Grade 6–76 marks

A student investigated the specific heat capacity of a 1.0 kg block of an unknown material.

She placed a 45 W heater and a thermometer in holes in the block and wrapped the block in insulation. She switched on the heater and recorded the temperature every 20 s.

The graph shows her results.

(a) Suggest why the temperature of the block rose more slowly during the first 80 s than it did later. [1]

(b) Use the graph to determine the specific heat capacity of the material.

Use readings at 100 s and 500 s. [4]

Suggest why. [1]

(a) the heater (itself) needed time to warm up (allow energy takes time to be conducted through the block to the thermometer)(1)
(b) energy transferred = 45 × 400 = 18 000 J(1)
(b) temperature rise = 48 − 24 = 24 °C (allow readings that give 23.8–24.2 °C)(1)
(b) c = 18 000 / (1.0 × 24) (allow ecf from their E and Δθ)(1)
(b) c = 750 J/kg °C (allow 740–760 J/kg °C)(1)
(c) to improve thermal contact between the thermometer and the block (allow so energy is transferred to the thermometer more easily / the thermometer reads the block's temperature more accurately; ignore 'to cool the thermometer')(1)
Answer: (a) heater takes time to warm up; (b) 750 J/kg °C; (c) better thermal contact
Examiner tip: In (b) the time is 400 s (500 − 100), not 500 s — use the time interval that matches your temperature change.

7Grade 4–55 marks

A storage heater contains a large block of bricks.

At night an electric heater raises the temperature of the bricks. During the day the bricks cool down and warm the room.

(a) The bricks are made from a material with a high specific heat capacity.

Explain why. [2]

(b) The mass of the bricks is 120 kg. During one day the temperature of the bricks falls from 300 °C to 40 °C.

specific heat capacity of brick = 840 J/kg °C

Calculate the energy transferred from the bricks. [3]

(a) (a high specific heat capacity means) more energy is stored for each kg for each °C rise in temperature (allow more energy needed to raise the temperature of 1 kg by 1 °C)(1)
(a) so more energy can be released to the room as the bricks cool (allow so the heater can keep the room warm for longer / a smaller mass of bricks is needed)(1)
(b) Δθ = 260 (°C)(1)
(b) ΔE = 120 × 840 × 260(1)
(b) ΔE = 26 208 000 J (allow 2.6 × 10⁷ J or 26 MJ; ignore a negative sign)(1)
Answer: (a) stores more energy per kg per °C, so releases more as it cools; (b) 26 208 000 J (2.6 × 10⁷ J)
Examiner tip: The temperature change is 300 − 40 = 260 °C — using 300 or 40 on its own is the most common lost mark.

8Grade 8–9Higher6 marks

A steel ball of mass 0.20 kg is heated to 95 °C. It is then dropped into an insulated cup containing 0.30 kg of water at 18 °C.

The ball and the water reach the same final temperature.

specific heat capacity of steel = 450 J/kg °C
specific heat capacity of water = 4200 J/kg °C

(a) Calculate the final temperature of the ball and the water.

Assume that no energy is transferred to the cup or to the surroundings.

Give your answer to 3 significant figures. [4]

(b) A student says:

"Energy is conserved, so the temperature of the ball falls by the same number of degrees as the temperature of the water rises."

Explain why the student is wrong. [2]

(a) energy transferred from the ball = energy transferred to the water (may be implied by an equation)(1)
(a) 0.20 × 450 × (95 − T) = 0.30 × 4200 × (T − 18)(1)
(a) correct rearrangement, e.g. 8550 + 22 680 = 1350 T or T = 31 230 / 1350(1)
(a) T = 23.1 °C (23.1 °C with no working scores 4 marks; 23 °C scores 3 marks)(1)
(b) the energy lost by the ball is equal to the energy gained by the water (ignore 'energy is conserved' alone)(1)
(b) but the water has a much greater mass and/or a much greater specific heat capacity, so its temperature change is much smaller (allow m × c for water = 1260 J/°C but for the ball only 90 J/°C / Δθ = ΔE/(mc) is different for each)(1)
Answer: (a) 23.1 °C (b) the same energy is transferred, but water has larger mass and SHC, so smaller temperature change
Examiner tip: Set up one equation with the unknown final temperature on both sides; the ball's change is (95 − T) and the water's is (T − 18).

9Grade 6–76 marks

A student used the apparatus in the diagram to determine the specific heat capacity of water.

She used 0.40 kg of water in a glass beaker with no lid. She did not stir the water. She switched the heater off after 10 minutes and immediately read the thermometer. She used the joulemeter reading and the temperature rise to calculate the specific heat capacity.

Her value was 5100 J/kg °C. The accepted value is 4200 J/kg °C.

Explain why her value was higher than the accepted value and how she could change the method to obtain a more accurate value. [6]

Calculation link: c = ΔE / (m Δθ), so a value that is too high means the energy reading is bigger than the energy gained by the water or the measured temperature rise is too small(1)
Energy is dissipated to the surroundings from the sides of the beaker / from the water surface (including by evaporation)(1)
Some energy heats the glass beaker (and the heater and thermometer) rather than the water(1)
Part of the heater may not be under the water, so energy is transferred to the air(1)
The water was not stirred, so the temperature at the thermometer may not be the average temperature of the water(1)
The temperature keeps rising after the heater is switched off (the hot heater still transfers energy), so reading it immediately gives a temperature rise that is too small(1)
Improvement: wrap the beaker in insulation (e.g. cotton wool / bubble wrap) and/or put a lid on the beaker to reduce energy dissipated(1)
Improvement: make sure the heater is fully submerged(1)
Improvement: stir the water before taking each reading(1)
Improvement: record the highest temperature reached after the heater is switched off(1)
Ignore: 'repeat and take a mean' unless linked to reducing random error; ignore 'human error'(1)

Level	Marks	What it looks like
Level 3	5–6	Gives a clear, logically linked explanation of why the measured value is too high, connecting at least two sources of error to their effect on the measured energy or temperature rise, and suggests matching improvements that would reduce each error.
Level 2	3–4	Identifies at least two relevant sources of error or improvements, with some explanation of how one of them affects the value, though links between error, effect and improvement may be incomplete.
Level 1	1–2	Makes one or two simple relevant points, such as energy being lost to the surroundings or insulating the beaker, with little or no explanation.
No relevant content	0	Nothing worthy of credit.

Answer: Energy dissipated to surroundings / heating the beaker and unstirred water / reading too early make ΔE too big or Δθ too small, so c = ΔE/(mΔθ) is too high; insulate, use a lid, submerge heater fully, stir, record the maximum temperature.
Examiner tip: Link every error to its effect on ΔE or Δθ and then to the value of c — a list of improvements with no explanation stays in Level 1–2.

10Grade 6–7Higher7 marks

An electric kettle has a power of 2.0 kW.

The kettle contains 0.80 kg of water at 20 °C. The kettle is switched on and the water is heated to 100 °C.

specific heat capacity of water = 4200 J/kg °C

(a) Calculate the energy transferred to the thermal energy store of the water. [2]

(b) Calculate the shortest possible time for the kettle to heat the water to 100 °C. [2]

Calculate the efficiency of the kettle. [3]

(a) Δθ = 80 (°C) seen or used(1)
(a) E = 268 800 J (allow 2.688 × 10⁵ J; allow 269 000 J or 270 000 J)(1)
(b) t = 268 800 / 2000 (allow ecf from (a); 2.0 kW must be converted to 2000 W)(1)
(b) t = 134 s (allow 134.4 s; allow 130 s; allow 2.2 minutes)(1)
(c) energy supplied = 2000 × 160 = 320 000 J(1)
(c) efficiency = 268 800 / 320 000 (allow ecf from (a); allow power method: useful power = 268 800 / 160 = 1680 W, then 1680 / 2000)(1)
(c) efficiency = 0.84 or 84% (do not accept a value greater than 1 or 100%)(1)
Answer: (a) 268 800 J (b) 134 s (c) 0.84 (84%)
Examiner tip: Convert 2.0 kW to 2000 W before using P = E/t, and remember efficiency compares useful energy out with total energy in, so it can never exceed 1.

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