A 1.50 kg block is heated by a 300 W heater. Its (time in s, temperature in °C) readings are (0, 20), (60, 31), (120, 42), (180, 53), (240, 64). (a) Find mean temperature-rise rate. (b) Assuming no loss, calculate specific heat capacity. (c) The accepted value is 820 J/(kg °C). Calculate mean energy-loss rate. (d) Explain two improvements and why the straight line cannot continue indefinitely.

Question

Accepted Answer

Rate 0.183 °C/s; uncorrected c = 1090 J/(kg °C); mean non-stored power about 74.5 W. Temperature rises 64 − 20 = 44 °C in 240 s, so mean rate is 44/240 = 0.183 °C/s. The equal eleven-degree rises each minute support approximate linearity over this interval. Heater supplies E = Pt = 300 × 240 = 72000 J. With no losses, c = E/(mΔT) = 72000/(1.50 × 44) = 1090 J/(kg °C). Using the accepted c, energy stored in block = 1.50 × 820 × 44 = 54120 J. Energy not stored = 72000 − 54120 = 17880 J, giving 17880/240 = 74.5 W. Insulate the block and improve heater/thermometer contact; record more points, fit a gradient and repeat. Include energy used to heat apparatus when refining the model. As temperature difference from the room grows, loss rate generally rises. Eventually heater power can balance losses, so temperature approaches a steady value instead of rising linearly forever.

Specific heat capacity from a heating curve

What a complete response should do

Block heating curve

Your self-assessment

Block heating curve — supplied values
Time / s	Temperature / °C
0	20
60	31
120	42
180	53
240	64