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  4. A hot body, obeying Newton's law of cooling is cooling down from its peak value 80°C to an ambient temperature of 30°C. It takes 5 minutes in cooling down from 80°C to 40°C. How much time will it take to cool down from 62°C to 32°C ? (Given In 2 = 0.693, In 5 = 1.609)

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Q. A hot body, obeying Newton's law of cooling is cooling down from its peak value 80$^{\circ}$C to an ambient temperature of 30$^{\circ}$C. It takes 5 minutes in cooling down from 80$^{\circ}$C to 40$^{\circ}$C. How much time will it take to cool down from 62$^{\circ}$C to 32$^{\circ}$C ?
(Given In 2 = 0.693, In 5 = 1.609)

JEE MainJEE Main 2014Thermal Properties of Matter

Solution:

$\frac{d\theta}{dt} = -C\left(\theta-\theta_{0}\right)$
$\int\limits^{40}_{80} \frac{1}{\theta-\theta_{0}} d\theta = -C/5$
$\int\limits^{32}_{62}\frac{1}{\theta -\theta _{0}} d\theta = -Ct$
$ln \left(\frac{80-30}{40-30}\right) = 5C$
$In \left(\frac{62-30}{40-50}\right) = Ct$
$In \,5 = 5c = 1.609$
$In \,16 = ct = 4 \times 0.693$
$t = 8.6 \,min$