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Q. A body takes $10$ minutes to cool down from $62^{\circ} C$ to $50^{\circ} C$. If the temperature of surrounding is $26^{\circ} C$ then in the next 10 minutes temperature of the body will be:

Rajasthan PMTRajasthan PMT 2002Thermal Properties of Matter

Solution:

From Newton's cooling law
$\frac{\theta_{1}-\theta_{2}}{t}=K\left(\frac{\theta_{1}+\theta_{2}}{2}-26\right)$
Case I : $\frac{12}{10}=K \times 30$
$ K =\frac{12}{10 \times 30} $
$=\frac{1}{25} $
Case II : $\frac{50-\theta_{2}}{10}=K\left(\frac{50+\theta_{2}}{2}-26\right)$
$\frac{\left(50-\theta_{2}\right)}{10}=\frac{1}{25}\left(\frac{\theta_{2}-2}{2}\right)$
$250-5\, \theta_{2}=\theta_{2}-2$
$ 6 \,\theta_{2} =252 $
$ \theta_{2} =\frac{252}{6}$
$=42^{\circ} C $