NEET Physics Question - Thermal Physics

A body cools from a temperature $3T$ to $2T$ in $10$ minutes. The room temperature is $T$. Assume that Newton's law of cooling is applicable. The temperature of the body at the end of next $10$ minutes will be: (2016 - II)

$\frac{4}{3}T$

$T$

$\frac{7}{4}T$

$\frac{3}{2}T$

Solution:

By Newton's law of cooling: $\frac{3T-2T}{10} = K(\frac{3T+2T}{2}-T) \Rightarrow \frac{T}{10} = K(1.5T)$. For next 10 mins: $\frac{2T-T'}{10} = K(\frac{2T+T'}{2}-T)$. Substituting $K = \frac{1}{15}$, we get $2T - T' = \frac{1}{15}(0.5T' + T) \times 10$. Solving gives $T' = \frac{3}{2}T$.

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