Ncert Exercises & Solutions

11.22:
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.

It is given in the question that:

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

According to Newton’s law of cooling,

$- \frac{dt}{dt} = K (T - T_{0}) \frac{dT}{K (T - T_{0})} = - Kdt . . . . . (i)$

Where,

The temperature of the body = T

K = constant

When the temperature of the body falls from 80°C to 50°C in time, t = 5 min = 300 s

Integrating equation (i),

$\int_{50}^{80} \frac{dT}{K (T - T_{0})} = - \int_{0}^{300} Kdt$

$\Rightarrow {[\log_{e} (T - T_{0})]}_{50}^{80} = - K {[t]}_{0}^{300} \Rightarrow \frac{2.3026}{K} \log_{10} (\frac{80 - 20}{50 - 20}) = - 300 \Rightarrow \frac{2.3026}{K} \log_{10} 2 = - 300 \Rightarrow K = \frac{- 2.3026}{300} \log_{10} 2$

Let the temperature of the body falls from 60°C to 30°C in time = t.
Hence,

$\frac{2.3026}{K} \log_{10} \frac{60 - 20}{30 - 20} = - t \frac{- 2.3026}{t} \log_{10} 4 = K . . . . (iii)$

Equating equations (ii) and (iii),

$\Rightarrow \frac{- 2.3026}{t} \log_{10} 4 = \frac{- 2.3026}{300} \log_{10} 2 \Rightarrow t = 300 \times 2 = 600 s = 10 \min$

Therefore, the time taken to cool the body from 60°C to 30°C is 10 minutes.

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