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A body takes 10 minutes to cool from $60^{\circ} \mathrm{C}$ to $50^{\circ} \mathrm{C}$. The temperature of surroundings is constant at $25^{\circ} \mathrm{C}$. Then, the temperature of the body after next 10 minutes will be approximately
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Verified Answer
The correct answer is:
$43^{\circ} \mathrm{C}$
$43^{\circ} \mathrm{C}$
According to Newton's law of cooling,
$$
\begin{aligned}
&\left(\frac{\theta_1-\theta_2}{t}\right)=K\left(\frac{\theta_1+\theta_2}{2}-\theta_0\right) \\
&\left(\frac{60-50}{10}\right)=K\left(\frac{60+50}{2}-25\right) \\
&\text { and, }\left(\frac{50-\theta}{10}\right)=K\left(\frac{50+\theta}{2}-25\right)
\end{aligned}
$$
Dividing eq. (i) by (ii),
$$
\frac{10}{(50-\theta)}=\frac{60}{\theta} \Rightarrow \theta=42.85^{\circ} \mathrm{C} \cong 43^{\circ} \mathrm{C}
$$
$$
\begin{aligned}
&\left(\frac{\theta_1-\theta_2}{t}\right)=K\left(\frac{\theta_1+\theta_2}{2}-\theta_0\right) \\
&\left(\frac{60-50}{10}\right)=K\left(\frac{60+50}{2}-25\right) \\
&\text { and, }\left(\frac{50-\theta}{10}\right)=K\left(\frac{50+\theta}{2}-25\right)
\end{aligned}
$$
Dividing eq. (i) by (ii),
$$
\frac{10}{(50-\theta)}=\frac{60}{\theta} \Rightarrow \theta=42.85^{\circ} \mathrm{C} \cong 43^{\circ} \mathrm{C}
$$
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