AQA · GCSE · Physics · Higher

Heating Energy of Compressed Air

Q06.3 6 marks

V = 5.0 \times 10^5\ \mathrm{m^3}

\rho = 48\ \mathrm{kg/m^3}

c = 1100\ \mathrm{J/(kg\,^{\circ}C)}

The temperature of the compressed air increases from 12 °C to 27 °C.

Calculate the energy transferred to the compressed air in the storage tank.

Use the Physics Equations Sheet.

Give your answer in standard form.

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Marking points

1 Uses density = mass/volume with 48 kg/m³ and 5.0 × 10⁵ m³.
2 Rearranges the density equation to multiply density by volume.
3 Obtains a compressed-air mass of 24 000 000 kg.
4 Uses ΔE = mcΔθ with c = 1100 J/kg °C and Δθ = 15 °C.
5 Calculates 396 000 000 000 J, allowing a consistent value from the calculated mass.
6 Expresses the result correctly in standard form as 3.96 × 10¹¹ J, allowing a consistent conversion.

Full-mark answer

First find the mass from density: $48 = m/(5.0 × 10^5)$, so $m = 48 \times 5.0 \times 10^5 = 2.4 \times 10^7\ \text{kg}$ . The temperature change is $27 - 12 = 15\,^{\circ}\text{C}$ . Then $δ E = mcδ θ = (2.4 × 10^7) × 1100 × 15 = 3.96 × 10^{11}\ text{J}$.

Why this answer loses marks

Uses the final temperature as the temperature change.

The energy equation needs the temperature rise, found by subtracting the starting temperature.

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