Physics 9702 · Paper 4

Measuring c

The experiment is simple — heat a known mass electrically and watch the temperature climb. The marks live in the graph method and in knowing which way errors push the answer.

The heated block

A metal block of known mass $m$ holds an electrical heater and a temperature probe. The heater's power comes from an ammeter and voltmeter: $P = IV$ .
Supply energy for a time $t$ and the books balance: $IVt = mc\,\Delta\theta$ . Everything is measured except $c$ .
Better than one reading: record temperature against time and take the gradient of the straight section. Then $P = mc \times \text{gradient}$ , and one bad reading cannot spoil the result.

Read the gradient, not one point: P = mc × gradient. Here 50 W heating 1.00 kg with gradient 16.4 °C per 400 s gives c = 50 / (1.00 × 0.041) ≈ 1200 J kg⁻¹ K⁻¹.

Worked example

Coursebook experiment: aluminium block

A 1.00 kg aluminium block is heated by a 12 V heater carrying 4.17 A. The graph of temperature against time is a straight line rising 16.4 °C every 400 s. Find $c$ .

Power in: $P = IV = 4.17 \times 12 = 50\ \text{W}$ .
Gradient: $\dfrac{\Delta\theta}{\Delta t} = \dfrac{16.4}{400} = 0.041\ \text{K s}^{-1}$ .
$c = \dfrac{P}{m \times \text{gradient}} = \dfrac{50}{1.00 \times 0.041} = 1220\ \text{J kg}^{-1}\text{K}^{-1}$ .

Answer

c \approx 1200\ \text{J kg}^{-1}\text{K}^{-1}

Error direction: why the result is too high

The accepted value for aluminium is about $900\ \text{J kg}^{-1}\text{K}^{-1}$ — the experiment above reads high. The reason: some heater energy escapes to the surroundings instead of warming the block, so each degree of rise seems to need more energy than it truly does. Energy loss always makes the measured $c$ too high.
Standard fixes, each worth a mark when asked: insulate the block; heat slowly so energy spreads through the block evenly; or the neat trick — start the block a few degrees below room temperature and stop the same amount above, so energy gained from the room early cancels energy lost to the room late.
When the substance is a liquid in a container, the container warms too. Either subtract the container's share $m_{\text{cup}} c_{\text{cup}} \Delta\theta$ from the energy first, or state that it was ignored (9702/41/M/J/23 Q3(c)(ii)).

Graph method: $P = mc \times \text{gradient}$ . Energy escaping makes measured $c$ too high; containers steal energy too.

Gases have two heat capacities

For a gas the answer depends on what the gas is allowed to do while heated. At constant volume all the energy raises the temperature; at constant pressure the gas also expands and does work pushing the surroundings back, so the same temperature rise needs more energy. The Thermodynamics chapter makes this exact with the first law — here just note that “the” specific heat capacity of a gas is only defined once the conditions are fixed.

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