Model temperature change with stated assumptions
Newton’s cooling model assumes the rate of temperature change is proportional to the difference from a constant ambient temperature. With time t, ambient temperature T_a and positive constant k, dT/dt = −k(T − T_a). The minus sign makes a hot object cool and a colder object warm toward the surroundings. The model is an approximation over conditions where k and the ambient temperature can reasonably be treated as constant.
Worked practice example
A component cools in a room at 20 °C. It is 80 °C at t = 0 and 50 °C after 10 minutes. Estimate its temperature after 20 minutes.
| Step | Working |
|---|---|
| General temperature model | T = 20 + Ce^(−kt). |
| Initial excess temperature | 80 − 20 = 60, so C = 60. |
| Use the second reading | 50 − 20 = 60e^(−10k), so e^(−10k) = 1/2 and k = ln(2)/10 per minute. |
| Predict at 20 minutes | T(20) = 20 + 60(1/2)² = 35 °C. |
Check your method
The excess temperature, not the total Celsius temperature, halves every 10 minutes. A prediction below 20 °C for this initially hot component would conflict with this model. Check that time units agree with the units of k.
Try it yourself
Using the same model, find the temperature after 30 minutes.
Show the answer and method
The excess is 60(1/2)³ = 7.5 °C, so T(30) = 27.5 °C. The temperature approaches 20 °C without crossing it in this ideal model.
This written practice example was prepared for Jemshah revision. It is additional practice, not an official examination question or a transcript of the video.