Hot water cools from $60^{\circ} C$ to $50^{\circ} C$ in the first 10 mins and to $42^{\circ} C$ in the next 10 mins. The temperature of surroundings

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$5^{\circ} C$

$10^{\circ} C$

$20^{\circ} C$

$40^{\circ} C$

answer is B.

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Detailed Solution

$G i v e n, T_{0} = 22^{\circ} C,$ From Newton's law of cooling, $\frac{T_{i} - T_{f}}{t} = (\frac{T_{i} + T_{f}}{2} - T_{0})$

For case-I, Hot soup cools from 60°C to 50°C in 10 min.

$\frac{60 - 50}{10} = - K [\frac{60 + 50}{2} - θ_{0}]$

For case-II, Time taken for cooling from $50 ° C$ to $42 ° C$

$\frac{50 - 42}{10} = - K [\frac{50 + 42}{2} - θ_{0}]$

After solving we get,

$\begin{array}{l} \frac{1}{0.8} = \frac{55 - θ_{0}}{46 - θ_{0}} \\ \Rightarrow \frac{10}{8} = \frac{55 - θ_{0}}{46 - θ_{0}} \\ \Rightarrow 460 - 10 θ_{0} = 440 - 8 θ_{0} \\ \Rightarrow 2 θ_{0} = 20 \\ \Rightarrow θ_{0} = 10^{\circ} C \end{array}$