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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Thermodynamics
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The temperature of a solid falls from $\;72^{0}C\;$ to $\;60^{0}C\;$ in 10 minutes . How much time will it take to cool from $\;60^{0}C\;$ to $\;52^{0}C\;$ if the temperature of surroundings is $\;36^{0}C\;$ c ?


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Answer : $\;15 \;min$
Explanation :
$\large\frac{dT}{dt} = K (\large\frac{\theta_{1}+\theta_{2}}{2} - \theta_{s})$
$\large\frac{12}{10} = K (66-36)$----(1)
$\large\frac{12}{t} = K (56 -36)$----(2)
$\large\frac{t}{10} = \large\frac{30}{20}$
$t = 15\;mins.$
answered Apr 8, 2014 by yamini.v

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