A body cools from a temperature to in 10 minutes. The room temperature is . Assume that Newton's law of cooling is applicable. The temperature of the body at the end of next 10 minutes will be [2016]

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Solution

Q.
A body cools from a temperature $3 T$ to $2 T$ 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]

1 $\frac{7}{4} T$

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

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

4 $T$

Ans.
(2)

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, $T_{s} = T,$ $T_{1} = \frac{3 T + 2 T}{2} = 2.5 T and \frac{d T_{1}}{d t} = \frac{3 T - 2 T}{10} = \frac{T}{10}$

$T_{2} = \frac{2 T + T'}{2} and \frac{d T_{2}}{d t} = \frac{2 T - T'}{10}$

So, $\frac{T}{10} = K (2.5 T - T)$ ...(i)

and $\frac{2 T - T'}{10} = K (\frac{2 T + T'}{2} - T)$ ...(ii)

Dividing eqn. (i) by eqn. (ii), we get

$\frac{T}{2 T - T'} = \frac{(2.5 T - T)}{(\frac{2 T + T'}{2} - T)} or \frac{2 T + T'}{2} - T = (2 T - T') \times \frac{3}{2}$

$T' = 3 (2 T - T') or, 4 T' = 6 T ∴ T' = \frac{3}{2} T$

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