Thermal energy transfer (specific heat and latent heat) – A-level Physics exam questions with mark scheme

1Core3 marks

An electric kettle has a power of 2.0 kW. It contains 0.50 kg of water at 18 °C.

Calculate the minimum time taken to heat the water to 100 °C. Specific heat capacity of water = 4200 J kg$^{-1}$ K$^{-1}$. [3]

$Q = mc\Delta\theta = 0.50 \times 4200 \times 82$ = 172 200 J (Δθ = 82 K seen)(1)
$t = Q/P = 172\,200 / 2000$(1)
$t$ = 86 s(1)
Answer: 86 s
Examiner tip: Use the temperature change (82 K), not the final temperature, and convert kW to W.

2Standard4 marks

A 0.040 kg ice cube at 0 °C is dropped into 0.30 kg of water at 25 °C in an insulated cup. All of the ice melts.

Calculate the final temperature of the water. Assume the cup has negligible heat capacity.

Specific heat capacity of water = 4200 J kg$^{-1}$ K$^{-1}$; specific latent heat of fusion of ice = 3.34 × 10$^5$ J kg$^{-1}$. [4]

energy to melt ice $= mL = 0.040 \times 3.34\times10^5$ = 13 400 J (13 360)(1)
energy lost by warm water = energy to melt ice + energy to warm melted ice: $0.30 \times 4200 \times (25 - T) = 13\,360 + 0.040 \times 4200 \times T$(1)
rearranges correctly, e.g. $1428\,T = 18\,140$(1)
$T$ = 13 °C (12.7)(1)
Answer: 13 °C (12.7 °C)
Examiner tip: Don't forget that the melted ice (now water at 0 °C) also has to warm up to the final temperature.

3Standard6 marks

A 50 W heater is used to heat a 0.20 kg sample of a solid in an insulated container. Assume that all the energy supplied by the heater is transferred to the sample. The graph shows how the temperature of the sample varies with time.

(a) Explain why the temperature stays constant between 120 s and 520 s. [2]

(b) Calculate the specific latent heat of fusion of the substance. [2]

(a) the substance is melting / changing state(1)
(a) energy supplied increases the potential energy of the particles (breaking bonds / increasing separation), not their kinetic energy, so temperature does not change allow 'energy is used to break bonds'; reject 'no energy is supplied'(1)
(b) energy supplied = $Pt = 50 \times 400$ = 20 000 J (time 400 s read from graph)(1)
(b) $L = \frac{20\,000}{0.20}$ = 1.0 × 10$^5$ J kg$^{-1}$(1)
(c) $c = \frac{Pt}{m\Delta\theta} = \frac{50 \times 120}{0.20 \times 60}$ (Δθ = 60 K)(1)
(c) $c$ = 500 J kg$^{-1}$ K$^{-1}$(1)
Answer: (b) 1.0 × 10$^5$ J kg$^{-1}$ (c) 500 J kg$^{-1}$ K$^{-1}$
Examiner tip: Read the duration of the flat section (520 − 120 = 400 s), not the time at which it ends.

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