Practical: investigate the specific heat capacity of a substance
4 marks · NormalA student takes repeated readings of temperature change in a metal block when it is heated for different lengths of time using a heater. The mass of the block is 0.4 kg.
| Heating time (s) | Temperature rise (°C) |
|---|---|
| 100 | 5.0 |
| 200 | 10.2 |
| 300 | 15.0 |
| 400 | 20.1 |
The heater supplies a constant power of 30 W throughout the experiment.
Plot a graph of temperature rise (y-axis) against heating time (x-axis) for this data. Draw a best-fit line.
Use your graph or the data to find the temperature rise after 250 seconds.
Calculate the specific heat capacity of the metal block using your value from (b).
Work this out on paper, then compare with the answer and worked solution below. With a free account you can submit written answers for marking.
Show answer
(a) Straight line graph: y-axis labelled 'Temperature rise (°C)', x-axis 'Heating time (s)', data points plotted, best-fit line drawn. · must show Correct axes and labels, Accurate plotting of all 4 points, Line of best fit through data
(b) About 12.7 °C (from interpolation between 10.2 °C at 200 s and 15.0 °C at 300 s). °C
(c) Q = 30 W × 250 s = 7500 J c = Q / (m × Δθ) = 7500 / (0.4 × 12.7) ≈ 148 J/kg °C
Must show: Graph axes correctly labelled and scaled, Best-fit straight line, Value interpolated for 250 s, Correct use of Q = m c Δθ
Common mistakes: Extrapolating not interpolating; Incorrect scale; Wrong substitution
Show mark scheme and worked solution
Marking points
B1 (a) · 1 mark · Plots all four points correctly and labels axes with units.
B2 (a) · 1 mark · Draws an appropriate best-fit straight line.
M1 (b) · 1 mark · Correctly finds (by interpolation) the temperature rise at 250 s.
A1 (c) · 1 mark · Correct substitution into c = Q/(m Δθ) and correct calculation.
Worked solution
a) Plot the graph with heating time (s) on the x-axis and temperature rise (°C) on the y-axis. Four points: (100, 5.0), (200, 10.2), (300, 15.0), (400, 20.1). Draw best-fit straight line.
b) Interpolate between 200 s (10.2 °C) and 300 s (15.0 °C): Gradient = (15.0 – 10.2)/(300 – 200) = 4.8/100 = 0.048 °C/s From 200 s to 250 s: 50 s further → temp rise: 10.2 + (0.048 × 50) = 10.2 + 2.4 = 12.6 °C (allow 12.7 °C rounding).
c) Q = P × t = 30 W × 250 s = 7500 J Specific heat capacity: c = Q / (m × Δθ) = 7500 / (0.4 × 12.7) = 7500 / 5.08 ≈ 147.6 ≈ 148 J/kg °C (2–3 s.f.)