In $5\, minutes,$ a body cools from $75^{\circ} {C}$ to $65^{\circ} {C}$ at room temperature of $25^{\circ} {C}$. The temperature of body at the end of next $5\, minutes$ is $......\,{ }^{\circ} {C} .$
JEE MAIN 2021, Medium
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By newton's law of cooling (with approximation)
$\frac{\Delta {T}}{\Delta {t}}=-{C}\left({T}_{{avg}}-{T}_{{s}}\right)$
$1^{\text {st }}\, \frac{-10^{\circ} {C}}{5\, {min}}=-{C}\left(70^{\circ} {C}-25^{\circ} {C}\right)$
${C}=\frac{2}{45}\, {min}^{-1}$
$2^{\text {nd }} \,\frac{{T}-65}{5 {min}}=-{C}\left(\frac{{T}+65}{2}-25\right)=-\left(\frac{2}{45}\right)\left(\frac{{T}+15}{2}\right)$
$9({T}-65)=-({T}+15)$
$10 {T}=570$
${T}=57^{\circ} {C}$
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