A body cools from 62,^oC to 50,^oC in 10, min | vedclass

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Question

Newton's law of cooling

using $\quad \frac{\Delta \mathrm{T}}{	ext { time }}=\frac{1}{\mathrm{K}}\left[\frac{	heta_{1}+	heta_{2}}{2}-	heta_

Vedclass · Accepted Answer

$26$

Vedclass · Answer

Newton's law of cooling

using $\quad \frac{\Delta \mathrm{T}}{	ext { time }}=\frac{1}{\mathrm{K}}\left[\frac{	heta_{1}+	heta_{2}}{2}-	heta_{0}ight]$

$\frac{62-50}{10}=\frac{1}{\mathrm{K}}\left[\frac{62+50}{2}-	hetaight]$

$\frac{50-42}{10}=\frac{1}{\mathrm{K}}\left[\frac{50+42}{2}-	hetaight]$

on dividing and solving

$	heta=26^{\circ} \mathrm{C}$

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 ........ $^oC$

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