BS Physics Lab: Thermal Conductivity and Heat Loss
A BS candidate measures the thermal conductivity of an aluminium alloy rod in a steady-state experiment. The rod is 300 mm long and 20 mm in diameter. A 30 W cartridge heater is clamped to one end; the other end sits in an ice-water bath at 0 °C. Two thermocouples are at x = 50 mm and x = 250 mm from the heater. At steady state they read 84.2 °C and 23.6 °C, while the room is 22 °C. The candidate calculates k = Q Δx / (A ΔT) using the heater power and the 200 mm spacing, obtaining about 315 W m⁻¹ K⁻¹. The handbook value for the alloy is 167 W m⁻¹ K⁻¹. The rod is wrapped in insulation, but a 5 mm gap exists near the heater, and the thermocouple beads are exposed. The candidate must decide whether the discrepancy is caused by heat loss, thermocouple calibration, or axial conduction into the clamp.
1. Calculate the apparent conductivity the candidate obtains and determine what effective heat flow between the thermocouples would be needed to recover the handbook value. What does this imply about the likely error source?
Read the worked answer
Area A = π(0.010)² = 3.14×10⁻⁴ m². The measured difference is 84.2 − 23.6 = 60.6 K over 0.20 m.
k_app = Q Δx / (A ΔT) = 30(0.20) / [(3.14×10⁻⁴)(60.6)] ≈ 315 W m⁻¹ K⁻¹.
For k = 167, the heat flow needed across the same segment is:
Q_eff = k A ΔT / Δx = 167(3.14×10⁻⁴)(60.6) / 0.20 ≈ 15.9 W.
Thus about 14 W of the 30 W must be lost before or between the thermocouples. That is large but plausible with an insulation gap and exposed beads. A common-mode thermocouple offset cancels in ΔT; explaining the error by calibration would require a gain error of roughly 47% in ΔT, which is unlikely. Heat loss is the leading explanation.
2. Describe a quick experiment using the existing thermocouples, heater, insulation, and ice bath that would distinguish heat loss from a thermocouple calibration error. State the predicted result for each case.
Read the worked answer
Repeat the run with the thermocouples swapped and then moved closer together, for example to x = 100 mm and x = 150 mm, keeping the same heater power and insulation. Compute local apparent k for two adjacent segments, 50–150 mm and 150–250 mm.
If heat loss dominates, the downstream segment gives a lower apparent k because less heat reaches it, and repairing the gap raises the apparent k. If one thermocouple has a calibration offset, swapping the probes changes the sign of the discrepancy, and an ice-point check reveals the offset. A common offset in both probes does not change ΔT, so it cannot explain a low k unless one probe is faulty.
3. The candidate proposes to fix the result by simply using the handbook value in the report. Explain why this is not acceptable and what a proper uncertainty statement should include.
Read the worked answer
Substituting the handbook value hides the measurement problem and is not a measurement. A proper report should give the measured k with an uncertainty budget: thermocouple resolution, spacing, rod diameter, heater power, and an estimated heat-loss correction of order 14 W.
A corrected value can be reported only after the heat-loss correction is measured, for example with a guard heater or by extrapolating local k to zero heat loss. The handbook value can be used as a comparison, not as a replacement for the measurement.