A body takes 10 minutes to cool down from 62^o C to 50^o C . If temperature of surro

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10-2.Transmission of Heat

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A body takes $10$ minutes to cool down from $62^o C$ to $50^o C$. If the temperature of surrounding is $26^o C$ then in the next $10$ minutes temperature of the body will be ......... $^oC$

A

$38$

B

$40$

C

$42$

D

$44$

Solution

From Newton's cooling law

$\frac{\theta_{1}-\theta_{2}}{t}=K\left(\frac{\theta_{1}+\theta_{2}}{2}-26\right)$

$Case-I:$ $\frac{12}{10}=\mathrm{K} \times 30$

$\mathrm{K}=\frac{12}{10 \times 30}=\frac{1}{25}$

$Case-II$ $: \frac{50-\theta_{2}}{10}=\mathrm{K}\left(\frac{50+\theta_{2}}{2}-26\right)$

$\frac{50-\theta_{2}}{10}=\frac{1}{25}\left(\frac{\theta_{2}-2}{2}\right)$

$250-5 \theta_{2}=\theta_{2}-2$

$6 \theta_{2}=252$

$\theta_{2}=\frac{252}{6}=42^{\circ} \mathrm{C}$

Standard 11

Physics

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