Answer to Question #153121 in Physics for Shehan Madushanka

Question

 A metal sphere, when suspended in a constant temperature enclosure, cools from 80 0C to 70 0C in 5 minutes and cool from 70 0C to 62 0C in the next five minutes. Calculate the temperature of the enclosure.  

Answer 1

Apply Newton's law of cooling, where $\theta_0$ is the temperature of surroundings:

$$\frac{\Delta\theta}{\Delta t}=k(\theta-\theta_0).$$

For the first case, we have

$$\frac{\theta_1-\theta_2}{\Delta t}=k\bigg(\frac{\theta_1+\theta_2}{2}-\theta_0\bigg),\\\space\\ \frac{80-70}{5}=k\bigg(\frac{80+70}{2}-\theta_0\bigg).$$

For the second case, by analogy

$$\frac{70-62}{5}=k\bigg(\frac{70+62}{2}-\theta_0\bigg).$$

Divide one by another:

$$\frac{80-70}{70-62}=\frac{75-\theta_0}{66-\theta_0},\\\space\\ \theta=30\text{°C}.$$