10-2.Transmission of Heat
medium

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

A

$43$

B

$47$

C

$41$

D

$45$

(JEE MAIN-2018)

Solution

According to $Newton's$ law of coolling,

$\left( {\frac{{{\theta _1} – {\theta _2}}}{t}} \right) = K\left( {\frac{{{\theta _1} + {\theta _2}}}{2} – {\theta _0}} \right)$

$\left( {\frac{{60 – 50}}{{10}}} \right) = K\left( {\frac{{60 + 50}}{2} – 25} \right)\,\,\,\,\,\,\,\,\,…\left( i \right)$

$and,\left( {\frac{{50 – \theta }}{{10}}} \right) = K\left( {\frac{{50 + \theta }}{2} – 25} \right)\,\,\,\,\,\,\,\,\,\,\,…\left( {ii} \right)$

Dividing eq. $(i)$ by $(ii)$,

$\frac{{10}}{{\left( {50 – \theta } \right)}} = \frac{{60}}{\theta } \Rightarrow \theta  = {42.85^ \circ }C \cong {43^ \circ }C$

Standard 11
Physics

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