NEET Physics Question - Thermal Physics

Certain quantity of water cools from $70^\circ\text{C}$ to $60^\circ\text{C}$ in the first $5$ minutes and to $54^\circ\text{C}$ in the next $5$ minutes. The temperature of the surroundings is: (2014)

$45^\circ\text{C}$

$20^\circ\text{C}$

$42^\circ\text{C}$

$10^\circ\text{C}$

Solution:

Using Newton's law of cooling: $\frac{70-60}{5} = K(65-T_s) \Rightarrow 2 = K(65-T_s)$ and $\frac{60-54}{5} = K(57-T_s) \Rightarrow 1.2 = K(57-T_s)$. Dividing gives $\frac{2}{1.2} = \frac{65-T_s}{57-T_s} \Rightarrow 5(57-T_s) = 3(65-T_s)$. Solving for $T_s$, we get $T_s = 45^\circ\text{C}$.

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