A covered beaker cools in a 20 °C room with (time in s, temperature in °C) readings (0, 80), (120, 68), (240, 59), (360, 52), (480, 47). (a) Find mean cooling rate over the first and final 240 s. (b) Explain the difference. (c) Design a fair comparison of two insulation materials, including controls, repeats and dependent variable. (d) Explain why equal thickness is not always the fairest engineering comparison.

Question

Accepted Answer

Rates 0.0875 °C/s and 0.0500 °C/s; cooling slows as temperature difference falls. Use matched containers and repeated, defined cooling measures. First 240 s: temperature drop is 80 − 59 = 21 °C, so mean rate = 21/240 = 0.0875 °C/s. Final 240 s: drop is 59 − 47 = 12 °C, giving 0.0500 °C/s. Thermal-transfer rate depends on temperature difference. The water–room difference falls from 60 °C toward 27 °C, reducing conduction, convection and net radiation rates. Use identical lidded containers with equal water mass and initial temperature, equal exposed area and simultaneous placement away from draughts. Change only insulation material. Use matched probes, record fixed intervals, compare time to a chosen temperature or drop over a chosen time, repeat and compare means and spread. Swap probes to check sensor bias. Equal thickness can mean unequal mass, cost or environmental burden because materials have different density. The controlled basis should match the design constraint: thickness, mass per area, price or available space.

Designing a fair insulation comparison

What a complete response should do

Cooling curve

Your self-assessment

Cooling curve — supplied values
Time / s	Temperature / °C
0	80
120	68
240	59
360	52
480	47