Question

A body takes 10 minutes to cool from 60C to 50C. The temperature of surroundings is constant at 25C. Then, the temperature of the body after next 10 minutes will be approximately

Answer 1

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From the Newton's law of cooling, we have

$t=\frac{1}{K} log_{e} (\frac{\theta_{2} - \theta_{0}}{\theta_{1} - \theta_{0}})$

Given,
For cooling a body from $60^{\circ}C$ to $50^{\circ}C$ , the time taken is 10 minutes

$t=\frac{1}{K} log_{e} (\frac{50 - 25}{60 - 25})$

$\implies t=\frac{1}{K}log_{e}(\frac{5}{7})$----[1]

Consider that, after another 10 min the temperature of the body falls to $\theta'$. 

Thus,
$t=\frac{1}{K} log_{e} (\frac{\theta' - 25}{50 - 25})$

$10=\frac{1}{K} log_{e} (\frac{\theta' - 25}{50 - 25})$

$10=\frac{1}{K} log_{e} (\frac{\theta' - 25}{25})$------[2]

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

$\frac{10}{10} = \frac{\frac{1}{K}log_{e}(\frac{5}{7})}{\frac{1}{K} log_{e} (\frac{\theta' - 25}{25})}$

$\implies \frac{5}{7} = \frac{\theta' - 25}{25}$

$\implies (\theta'- 25) = \frac{25×5}{7}$

$\implies \theta'= \frac{125}{7}+25$

$\implies \theta'= \frac{125}{7}+25$

$\implies \theta'= \frac{300}{7}$

$\implies \theta'= 42\frac{6}{7}$

$\implies \theta'=42.85^{\circ}C$

Therefore,
The temperature of the body after next 10 minutes will be approximately $42.85^{\circ}C$