10-2.Transmission of Heat
medium

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 loge $log_e(\theta - \theta_0) $ and $t$ is

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(AIEEE-2012)

Solution

$Newton's\,law$ of cooling

$\frac{d \theta}{d t}=-k\left(\theta-\theta_{0}\right)$

$\int_{\theta_{1}}^{\theta} \frac{d \theta}{\theta-\theta_{0}}=-\int_{0}^{t} k d t$

$\ln \left(\theta-\theta_{0}\right)-\ln \theta_{1}=-k t$

$\ln \left(\theta-\theta_{0}\right)=-k t+\ln \theta_{1}$

Standard 11
Physics

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