A body cools from 62,^oC to 50,^oC in 10, minutes and to 42,^oC is next 10, minutes . temp. of surro

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10-2.Transmission of Heat

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A body cools from $62\,^oC$ to $50\,^oC$ in $10\, minutes$ and to $42\,^oC$ is next $10\, minutes$. The temp. of surrounding is ........ $^oC$

A

$16$

B

$26$

C

$36$

D

$21$

Solution

Newton's law of cooling

using $\quad \frac{\Delta \mathrm{T}}{\text { time }}=\frac{1}{\mathrm{K}}\left[\frac{\theta_{1}+\theta_{2}}{2}-\theta_{0}\right]$

$\frac{62-50}{10}=\frac{1}{\mathrm{K}}\left[\frac{62+50}{2}-\theta\right]$

$\frac{50-42}{10}=\frac{1}{\mathrm{K}}\left[\frac{50+42}{2}-\theta\right]$

on dividing and solving

$\theta=26^{\circ} \mathrm{C}$

Standard 11

Physics

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