Thermal Properties of Matter - Result Question 47
48. Certain quantity of water cools from $70^{\circ} C$ to $60^{\circ} C$ in the first 5 minutes and to $54^{\circ} C$ in the next 5 minutes. The temperature of the surroundings is:
[2014]
(a) $45^{\circ} C$
(b) $20^{\circ} C$
(c) $42^{\circ} C$
(d) $10^{\circ} C$
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Answer:
Correct Answer: 48. (a)
Solution:
- (a) Let the temperature of surroundings be $\theta_0$ By Newton’s law of cooling
$ \begin{align*} & \frac{\theta_1-\theta_2}{t}=k[\frac{\theta_1+\theta_2}{2}-\theta_0] \\ \Rightarrow & \frac{70-60}{5}=k[\frac{70+60}{2}-\theta_0] \\ \Rightarrow & 2=k[65-\theta_0] \tag{i} \end{align*} $
Similarly, $\frac{60-54}{5}=k[\frac{60+54}{2}-\theta_0]$
$\Rightarrow \frac{6}{5}=k[57-\theta_0]$
By dividing (i) by (ii) we have
$ \frac{10}{6}=\frac{65-\theta_0}{57-\theta_0} \Rightarrow \theta_0=45^{\circ} $
