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એક ગરમ પદાર્થ ન્યુટનના નિયમનું પાલન કરીને તેના મહત્તમ તાપમાન $80\,^oC$ થી ઠંડો પાડીને વાતાવરણનું તાપમાન $30\,^oC$ પર આવે છે.તાપમાન $80\,^oC$ થી $40\,^oC$ થતાં $5\, minutes$ લાગે છે તો $62\,^oC$ થી $32\,^oC$ થતાં .......... $\min.$ લાગે? ($ln\, 2\, = 0.693, ln\, 5\, = 1.609$)
$3.75$
$8.6$
$9.6$
$6.5$
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 ${62^ \circ }C\,to\,{32^ \circ }C,$
$t=8.6\,minutes.$
