The method and the results are on the specific heat capacity lab. Do the analysis yourself first, then check the answers below.
Quick answer: Specific heat capacity is the energy needed to warm 1 kg of a material by 1 °C, and energy transferred equals m × c × Δθ. These answers go with the Physics 9-10 lab on specific heat capacity: the graph, the calculations, the uncertainties and the conclusion explained, for all 8 analysis questions.
Model answers use the sample data. Your own values will differ a little, but the specific heat capacity should lie between 900 and 1100 J/(kg °C) with decent insulation. Always compare with the book value and explain the sign of the error.
Level: Physics grades 9-10 (GCSE · IGCSE · O Level · Class 9-10) · Curriculum: GCSE P3, IGCSE 2.2
The gradient is ΔE / Δθ = m c, so c = gradient / m. Gradient = (28 800 − 0) / (50.2 − 20.0) = 28 800 / 30.2 = 954 J/°C, giving c ≈ 950 J/(kg °C).
Question 5 Comparing with the data book
(954 − 900) / 900 × 100 = 6.0 %, so the result is about 6 % too high.
Question 6 Why the value is too high
Some energy escapes from the block to the air, so the block gains less energy than the 48 W supplied. The calculation assumes all the energy was gained, so it divides by a temperature rise that is too small and gives a value too large. Insulation reduces the loss and brings the result closer to 900 J/(kg °C).
Question 7 Uncertainty
The rise uses two readings, so the uncertainty is about ±1.0 °C. 1.0 / 30.2 × 100 ≈ 3.3 %.
Question 8 Improvements
Wrap the block in thicker insulation (and put a lid on top) to reduce losses; continue heating for a few minutes after switching off to find the true maximum; use a joulemeter for the energy; repeat and average.
