physics ncert solutions class 11th Physics Class 11th Thermal Properties of Matter

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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5 months ago

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 temperature of the body falls from 60 $?$ to 30 $? i n t i m e = t'$

Hence, we get:

$\frac{2.3026}{K} \log_{10} ? \frac{60 - 20}{30 - 20}$ = -t'

$? \frac{- 2.3026}{t'} \log_{10} ? 4$ = K.(iii)

Equating equations (ii) and (iii), we get,

$\frac{2.3026}{300} \log_{10} ? 2$ = $\frac{- 2.3026}{t'} \log_{10} ? 4$

$? t^{'} = 300 \times 2 = 600 s$ = 10 min

Hence, the time taken to cool the body from 60 $?$ to 30 $?$ is 10 minutes.

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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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