11.20 A body cools from 80 °C to 50 °C in 5 minutes. Calculate the time it takes to cool from 60 °C to 30 °C. The temperature of the surroundings is 20 °C.

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11.20 According to Newtonâ€™s law of cooling, we have $- \frac{d T}{d t}$
= K(T - $T_{o}$ ) or $\frac{d T}{K (T - T_{o})}$ = -Kdtâ€¦â€¦(i)Where, temperature of the body = T

Temperature of the surroundings = $T_{o}$ = 20 °C

K is a constant

The temperature of the body falls from 80 $?$ to 50 $? i n t i m e t = 5 \min ? = 300 s$

Integrating equation (i), we get

$5080 \frac{d T}{K (T - T_{o})}$ = - $\int_{0}^{300} K d t∫$

$? {[\log_{e} ? (T - T_{o})]}_{30}^{80}$ = -K ${[t]}_{0}^{300}$

$? \frac{2.3026}{K} \log_{10} ? \frac{80 - 20}{50 - 20}$ = -300

$? \frac{2.3026}{300} \log_{10} ? 2$ = K â€¦.(ii)

The temperatur

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