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10-2.Transmission of Heat
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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
