Heating a Block: Specific Heat and Energy Transfer
A 2.0 kg aluminium block is heated by an electric heater rated at 48 W. The block's temperature rises from 20.0 °C to 45.0 °C in 12.0 minutes. The specific heat capacity of aluminium is 900 J/(kg·K). A student calculates the energy supplied by the heater and compares it with the energy gained by the block. The block is not insulated and sits in a room at 20.0 °C. A second trial uses a 1.0 kg block and the same heater, and the temperature rise in the same time is larger.
1. Calculate the energy supplied by the heater and the energy gained by the block, and determine the efficiency of the heating process.
Read the worked answer
Energy supplied: E_in = P t = 48 × (12.0 × 60) = 48 × 720 = 34 560 J.
Energy gained by block: E_gain = mcΔT = 2.0 × 900 × 25.0 = 45 000 J.
The calculated gain (45 000 J) is larger than the energy supplied (34 560 J), which is impossible for a passive heater. The apparent efficiency would be 45 000 / 34 560 = 1.30, i.e. 130%.
This signals an error or a missing factor. Possibilities: the heater power is higher than 48 W, the time is longer than 12.0 min, the mass is greater than 2.0 kg, or the specific heat value used is too high. A good answer should not simply report 130% as an efficiency; it should identify the inconsistency and propose which measurement to check. The most likely culprit is the stated heater power, since 48 W is modest for this mass.
2. Explain why the second trial, with a 1.0 kg block, shows a larger temperature rise, and state what this shows about the relationship between mass and temperature change.
Read the worked answer
For the same energy input and the same material, ΔT = E / (mc). Halving the mass doubles the temperature change (assuming the same efficiency and the same losses).
So a 1.0 kg block should rise by about 50 °C in the same time if the energy transfer is unchanged, rather than 25 °C.
This shows that temperature change is inversely proportional to mass for a given energy input and material. It also shows that specific heat capacity is an intensive property: it does not depend on how much aluminium is present, but the temperature response does.
A subtlety: the smaller block has a larger surface-area-to-mass ratio, so it loses heat faster per kilogram. The observed rise may therefore be slightly less than double, and the comparison is a useful check on how significant heat loss is.
3. Design a modification that would let you measure the specific heat capacity more reliably, and explain how it addresses the main problem in the original setup.
Read the worked answer
Insulate the block, for example by wrapping it in foam and placing it in a lid-covered container, and use a thermometer embedded in a drilled hole so the temperature is measured inside the metal rather than at the surface.
The main problem is heat loss to the room, which makes the energy gained by the block less than the energy supplied and biases the calculated specific heat upward (or, in this case, creates an impossible energy budget). Insulation reduces the loss; measuring the block's temperature rather than the air or surface avoids a systematic error.
A further improvement is to plot temperature against time and use the initial gradient, before losses become large, or to correct for cooling by recording the cooling curve after the heater is switched off and extrapolating. Repeating with different masses and plotting E against mΔT gives a straight line whose gradient is c, averaging random errors.