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9702 · 14.3

Specific heat capacity and specific latent heat — practice questions

Practice and worked examples for 9702 Specific heat capacity and specific latent heat. Short previews only — attempt the full question in MarkScheme against the official scheme.

Worked example 1

A 2.5 kg block of aluminium is heated, absorbing 120 kJ of energy. If its temperature rises from 20 °C to 70 °C, calculate the specific heat capacity of aluminium.

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  1. Identify knowns: m=2.5m = 2.5 kg, Q=120Q = 120 kJ =120,000= 120,000 J, ΔT=70−20=50\Delta T = 70 - 20 = 50 K (or °C).
  2. Recall the formula: Q=mcΔTQ = mc\Delta T.
  3. Rearrange for cc: c=QmΔTc = \frac{Q}{m\Delta T}.
  4. Substitute values: c=120,0002.5×50c = \frac{120,000}{2.5 \times 50}.
  5. Calculate: c=120,000125=960c = \frac{120,000}{125} = 960 J kg⁻¹ K⁻¹.

Worked example 2

An immersion heater rated at 50 W is used to melt 0.030 kg of ice at 0 °C in 2 minutes. Assuming no heat loss, calculate the specific latent heat of fusion of ice.

Show solution outline
  1. Identify knowns: Power P=50P = 50 W, mass m=0.030m = 0.030 kg, time t=2t = 2 min =120= 120 s.
  2. Calculate energy supplied: Q=P×t=50×120=6000Q = P \times t = 50 \times 120 = 6000 J.
  3. Recall the formula for latent heat: Q=mLQ = mL.
  4. Rearrange for LL: L=QmL = \frac{Q}{m}.
  5. Substitute values: L=60000.030L = \frac{6000}{0.030}.
  6. Calculate: L=200,000L = 200,000 J kg⁻¹.
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