If a liquid takes $30 \;sec$ in cooling from $80^{\circ} C$ to $70^{\circ} C$ and $70 \;sec$ in cooling from $60^{\circ} C$ to $50^{\circ} C$, then find the room temperature.
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$\frac{\theta_1-\theta_2}{t}=K\left(\frac{\theta_1+\theta_2}{2}-\theta_0\right)$
In the first case,
$\frac{80-70}{3} 0=K\left(\frac{80+70}{2}-\theta_0\right), \frac{1}{3}=K\left(75-\theta_0\right)$
In the second case,
$\frac{60-50}{70}=K\left(\frac{60+50}{2}-\theta_0\right), \frac{1}{7}=K\left(55-\theta_0\right)$
$\frac{7}{3}=\frac{\left(75-\theta_0\right)}{\left(55-\theta_0\right)} \Rightarrow 385-7 \theta_0=225-3 \theta_0$
$\Rightarrow \theta_0=\frac{160}{4}=40^{\circ} C$
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