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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 $\log _e\left(\theta-\theta_0\right)$ and $t$ is
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According to Newtons law of cooling.
$\frac{\mathrm{d} \theta}{\mathrm{dt}} \propto-\left(\theta-\theta_0\right)$
$\Rightarrow \quad \frac{\mathrm{d} \theta}{\mathrm{dt}}=-\mathrm{k}\left(\theta-\theta_0\right)$
$\int \frac{d \theta}{\theta-\theta_0}=\int-k d t$
$\Rightarrow \quad \ln \left(\theta-\theta_0\right)=-k t+c$
Hence the plot of $\ln \left(\theta-\theta_0\right)$ vs $t$ should be a straight line with negative slope.
$\frac{\mathrm{d} \theta}{\mathrm{dt}} \propto-\left(\theta-\theta_0\right)$
$\Rightarrow \quad \frac{\mathrm{d} \theta}{\mathrm{dt}}=-\mathrm{k}\left(\theta-\theta_0\right)$
$\int \frac{d \theta}{\theta-\theta_0}=\int-k d t$
$\Rightarrow \quad \ln \left(\theta-\theta_0\right)=-k t+c$
Hence the plot of $\ln \left(\theta-\theta_0\right)$ vs $t$ should be a straight line with negative slope.
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