10-2.Transmission of Heat
hard

A hot body, obeying Newton's law of cooling is cooling down from its peak value $80\,^oC$ to an ambient temperature of $30\,^oC$ . It takes $5\, minutes$ in cooling down from $80\,^oC$ to $40\,^oC$.  ........  $\min.$ will it take to cool down from $62\,^oC$ to $32\,^oC$ ? (Given $ln\, 2\, = 0.693, ln\, 5\, = 1.609$)

A

$3.75$

B

$8.6$

C

$9.6$

D

$6.5$

(JEE MAIN-2014)

Solution

From $Newton's\,law$ of cooling,

$t = \frac{1}{k}{\log _e}\left( {\frac{{{\theta _2} – {\theta _0}}}{{{\theta _1} – {\theta _0}}}} \right)$

From question and above equation,

$5 = \frac{1}{k}{\log _e}\frac{{\left( {40 – 30} \right)}}{{\left( {80 – 130} \right)}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,…\left( 1 \right)$

$And,t = \frac{1}{k}{\log _e}\frac{{\left( {32 – 30} \right)}}{{\left( {62 – 30} \right)}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,…\left( 2 \right)$

Dividing equation $(2)$ by $(1)$,

$\frac{t}{5} = \frac{{\frac{1}{k}{{\log }_e}\frac{{\left( {32 – 30} \right)}}{{\left( {62 – 30} \right)}}}}{{\frac{1}{k}{{\log }_e}\frac{{\left( {40 – 30} \right)}}{{\left( {80 – 30} \right)}}}}$

On solving we get, time taken to cool down from ${62^ \circ }C\,to\,{32^ \circ }C,$

$t=8.6\,minutes.$

Standard 11
Physics

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