The Specific Heat That Keeps Coming Out Too High
To measure the specific heat capacity of aluminium, a student uses a 1.00 kg block with two drilled holes, one for a 30 W heater (12 V, 2.5 A) and one for a thermometer. Run 1: starting at 20.0 °C, the heater runs for 300 s; the thermometer reads 28.5 °C at switch-off and peaks at 28.9 °C half a minute later, giving c = 9000/(1.00 × 8.5) ≈ 1060 J kg⁻¹ K⁻¹. Run 2: starting again at 20.0 °C but heating for 600 s, the final reading is 35.6 °C, giving about 1150 J kg⁻¹ K⁻¹. The accepted value for aluminium is about 900 J kg⁻¹ K⁻¹ and the room stays at 20.0 °C. In a separate check the hot block cools from 28.9 °C to 25.0 °C in 12 minutes with the heater off. The student argues that the two runs bracket the accepted value, so their average must be reliable.
1. Show that the two runs disagree in a way that is not random, identify the systematic error and its direction, and explain why the longer run is worse.
Read the worked answer
The two values. Run 1: c = Pt/(mΔT) = (30 × 300)/(1.00 × 8.5) = 9000/8.5 = 1.06 × 10³ J kg⁻¹ K⁻¹. Run 2: c = (30 × 600)/(1.00 × 15.6) = 18 000/15.6 = 1.15 × 10³ J kg⁻¹ K⁻¹.
Both exceed 900, and the hotter run exceeds it by more. That pattern is the signature of a systematic error, not random scatter. The block loses heat to the room throughout, so the energy actually stored is Pt − Q_lost, and
c_true = (Pt − Q_lost)/(mΔT) < Pt/(mΔT).
The student's formula therefore always overestimates c. The hotter the block, the larger the excess temperature above the room and the larger the loss rate (Newton's law of cooling) - exactly why run 2 is worse. Two numbers biased in the same direction cannot be averaged away; the mean of 1060 and 1150 is still far too high.
The 0.4 K after switch-off. The heater is hotter than the block when the current stops, and the heat stored in it continues to flow into the block. Using 28.5 °C instead of 28.9 °C underestimates ΔT and inflates c again. That extra 0.4 K was produced by energy supplied during the 300 s, so the equilibrium reading of 28.9 °C is the one that belongs with Pt.
2. Use the cooling data to estimate the heat lost during run 1 and obtain a corrected value of c. State your assumptions and comment on how close the answer is to 900 J kg⁻¹ K⁻¹.
Read the worked answer
Loss coefficient from the cooling check. The block falls from 28.9 °C to 25.0 °C in 720 s, so the mean temperature during cooling is 26.95 °C, an excess of 6.95 K above the room. The mean cooling rate is 3.9/720 = 5.4 × 10⁻³ K s⁻¹, giving a mean loss rate of
P = mc(dT/dt) ≈ 1.00 × 900 × 5.4 × 10⁻³ ≈ 4.9 W.
If loss is proportional to the excess temperature, h = 4.9/6.95 ≈ 0.70 W K⁻¹.
Loss during run 1. The excess temperature climbs from 0 to 8.9 K, so the mean excess is about 4.45 K and the mean loss rate is 0.70 × 4.45 ≈ 3.1 W. Over 300 s, Q_lost ≈ 3.1 × 300 ≈ 930 J.
Corrected value. Using the equilibrium rise of 8.9 K (not 8.5 K), which includes the heat still arriving from the heater after switch-off:
c = (9000 − 930)/(1.00 × 8.9) = 8070/8.9 ≈ 907 J kg⁻¹ K⁻¹.
Assumptions. Newton's law of cooling holds over this small range; the block is effectively isothermal because aluminium conducts well; the temperature rises linearly with time, so the arithmetic mean excess is fair; the heat capacities of the heater, thermometer and drilled holes are neglected, which would push the answer down slightly. Assuming instead ΔT = 8.5 K gives about 950 J kg⁻¹ K⁻¹, so the correction brings the result from 18% high to within about 1–5% of 900 - good evidence that heat loss was indeed the dominant error.
3. Suggest two practical improvements and explain what the initial-gradient method would avoid.
Read the worked answer
Improvements. (i) Lag the block in foam or felt and cap the holes; this cuts h sharply and makes the correction small. (ii) Keep the temperature rise to a few kelvin, by using a lower-power heater or a shorter run, so that the mean excess temperature - and hence the loss - stays small. (iii) Record the temperature continuously with a data logger, so the answer comes from a fitted graph rather than from a single pair of readings. (iv) If the heater is heavy, include its heat capacity: c_block = (Pt − m_h c_h ΔT)/(m_block ΔT).
What the initial-gradient method avoids. Plot temperature against time and take the gradient at t = 0, when the block is exactly at room temperature. At that instant the excess temperature is zero, so by Newton's law the heat loss is also zero, and
c = P/(m × gradient at t = 0)
needs no cooling correction at all. It also disposes of the thermal-lag problem, because the gradient is measured while the heater is still on and the heater and block are close to the same temperature - using the post-switch-off peak is what introduced the awkward 8.5 versus 8.9 K choice. Its weakness is that the initial gradient is hard to measure precisely: the first few seconds are noisy, and a straight-line fit over the first 30 s of logged data is the usual compromise. Note that neither method fixes a mis-calibrated thermometer, so the ice-point and steam-point checks should still be done.