Answer to Question #259643 in Physics for Casper b

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.why did you divide the temperatures by two

Answer 1

Below $\theta_0$ is the temperature of the enclosure.

By Newton's law of cooling, during the cooling from 80 to 70°C, where $\theta_1=80°\text C,\theta_2=70°\text C,\text dt=5\text{ min}$:

$\frac{\text d\theta}{\text dt}=C(\theta-\theta_0),\\\space\\ \frac{\theta_1-\theta_2}{\text dt}=C\bigg(\frac{\theta_1+\theta_2}{2}-\theta_0\bigg),\\\space\\ 2=C(75-\theta_0).$

Cooling from 70 to 62°, where $\theta_1=70°\text C,\theta_2=62°\text C,\text dt=5\text{ min}$:

$\frac{\text d\theta}{\text dt}=C(\theta-\theta_0),\\\space\\ \frac{\theta_1-\theta_2}{\text dt}=C\bigg(\frac{\theta_1+\theta_2}{2}-\theta_0\bigg),\\\space\\ 1.6=C(66-\theta_0).$

Divide one equation by another:

$\frac 2{1.6}=\frac{C(75-\theta_0)}{C(66-\theta_0)},\\\space\\ \theta_0=30°\text C.$

In this problem, we put

$\theta=\frac{\theta_1+\theta_2}{2}$

because we need to use the average temperature of the system during the given interval.