First slide

Radiation

Question

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

Moderate

A

$\frac{7}{4} T$
B

$\frac{3}{2} T$
C

$\frac{4}{3} T$
D

T

Solution

According to Newton's law of cooling,
$\frac{d T}{d t} = K (T - T_{s})$
$For two cases, \frac{d T_{1}}{d t} = K (T_{1} - T_{s}) and \frac{d T_{2}}{d t} = K (T_{2} - T_{s})$
$Here, T_{s} = T, T_{1} = \frac{3 T + 2 T}{2} = 2.5 T$
$\begin{array}{l} and \frac{d T_{1}}{d t} = \frac{3 T - 2 T}{10} = \frac{T}{10} \\ T_{2} = \frac{2 T + T^{'}}{2} and \frac{d T_{2}}{d t} = \frac{2 T - T^{'}}{10} \end{array}$
$So, \frac{T}{10} = K (2.5 T - T)$ ………….(1)
$\frac{2 T - T^{'}}{10} = K (\frac{2 T + T^{'}}{2} - T)$ ………(2)
From eq (1) & (2)
$\begin{array}{l} \frac{T}{2 T - T^{'}} = \frac{(2.5 T - T)}{(\frac{2 T + T^{'}}{2} - T)} \\ \frac{2 T + T^{'}}{2} - T = (2 T - T^{'}) \times \frac{3}{2} \\ T^{'} = 3 (2 T - T^{'}) or, 4 T^{'} = 6 T \\ ∴ T^{'} = \frac{3}{2} T \end{array}$

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