Question

A liquid in a beaker has temperature $\theta(t)$ at time $t$ and $\theta_0$ is temperature of surroundings, then according to Newton's law of cooling the correct graph between $log_{e}\left(\theta-\theta_{0}\right)$ and $t$ is :

• A.
• B.
• C.
• D.

Answer 1

Answer: (A)

Answer 2

$\frac{d\theta}{dt}=-k\left(\theta-\theta_{0}\right)$ $\int\limits^\theta_{{\theta_0}}$ $\frac{d\theta}{\theta-\theta_{0}}=-k\int\limits^t_{{0}}dt$ $In \left(\theta-\theta_{0}\right) = -kt + C$ So graph is straight line.