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 ....... $^oC$
JEE MAIN 2018, Medium
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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 $(ii)$,
$\frac{{10}}{{\left( {50 - \theta } \right)}} = \frac{{60}}{\theta } \Rightarrow \theta = {42.85^ \circ }C \cong {43^ \circ }C$
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