A body cools down from $52.5^{\circ} \mathrm{C}$ to $47.5^{\circ} \mathrm{C}$ in $5 \mathrm{~min}$ and from…
A body cools down from $52.5^{\circ} \mathrm{C}$ to $47.5^{\circ} \mathrm{C}$ in $5 \mathrm{~min}$ and from $47.5^{\circ} \mathrm{C}$ to $42.5^{\circ} \mathrm{C}$ in $7.5 \mathrm{~min}$. Then, the temperature of the surroundings is
$39^{\circ} \mathrm{C}$
$25^{\circ} \mathrm{C}$
$35^{\circ} \mathrm{C}$
$15^{\circ} \mathrm{C}$
Solution
Given that, for first condition,
Initial temperature, $T_1=52.5^{\circ} \mathrm{C}$
Final temperature, $T_2=47.5^{\circ} \mathrm{C}$
Average temperature, $T_{\mathrm{av}}=\frac{T_1+T_2}{2}$
$
=\frac{525+47.5}{2}=50^{\circ} \mathrm{C}
$
For second condition,
Initial temperature, $T_1=47.5^{\circ} \mathrm{C}$
Final temperature, $T_2=42.5^{\circ} \mathrm{C}$
Average temperature, $T_{\mathrm{av}}=\frac{T_1+T_2}{2}$
$
=\frac{47.5+42.5}{2}=45^{\circ} \mathrm{C}
$
Let $T_0$ be the temperature of surroundings. Now, using expression of rate of cooling,
$
R=\frac{\Delta T}{\Delta t}=-K\left(T_{\mathrm{av}}-T_0\right)
$
Substituting the values in above equation from lst and 2 nd conditions, we get
$
\frac{T_1-T_2}{t_1}=-K\left(T_{\mathrm{av}}-T_0\right)
$
$
\frac{52.5-47.5}{5}=-K\left(50-T_0\right)...(i)
$
Similarly, $\frac{47.5-42.5}{7.5}=-K\left(45-T_0\right)...(ii)$
Dividing Eq. (i) by Eq. (ii), we get
$
\begin{aligned}
\frac{\frac{52.5-47.5}{5}}{\frac{47.5-42.5}{7.5}} & =\frac{-K\left(50-T_0\right)}{-K\left(45-T_0\right)} \\
\Rightarrow \quad \frac{5}{5} \times \frac{7.5}{5} & =\frac{50-T_0}{45-T_0} \Rightarrow \frac{3}{2}=\frac{50-T_0}{45-T_0} \\
T_0 & =35^{\circ} \mathrm{C}
\end{aligned}
$
Hence, the temperature of surroundings is $35^{\circ} \mathrm{C}$