A body cools from $80\,^{\circ} C$ to $50\,^{\circ} C$ in $5$ minutes. Calculate the time (in $min$) it takes to cool from $60\,^{\circ} C$ to $30\,^{\circ} C .$ The temperature of the surroundings is $20\,^{\circ} C$
A body cools from $80\,^{\circ} C$ to $50\,^{\circ} C$ in $5$ minutes. Calculate the time (in $min$) it takes to cool from $60\,^{\circ} C$ to $30\,^{\circ} C .$ The temperature of the surroundings is $20\,^{\circ} C$
According to Newton's law of cooling, we have:
$-\frac{d T}{d t}=K\left(T-T_{0}\right)$
$\frac{d T}{K\left(T-T_{0}\right)}=-K d t...(i)$
Where,
Temperature of the body $=T$
Temperature of the surroundings $=T_{0}=20^{\circ} C$
$K$ is a constant
Temperature of the body falls from $80^{\circ} C$ to $50^{\circ} C$ in time, $t=5 min =300 s$
Integrating equation $(i),$ we get
$\int_{00}^{80} \frac{d T}{K\left(T-T_{0}\right)}=-\int_{0}^{300} K d t$
$\left[\log _{e}\left(T-T_{0}\right)\right]_{s_{0}}^{80}=-K[t]_{0}^{300}$
$\frac{2.3026}{K} \log _{10} \frac{80-20}{50-20}=-300$
$\frac{2.3026}{K} \log _{10} 2=-300$
$\frac{-2.3026}{300} \log _{10} 2=K\dots (ii)$
The temperature of the body falls from $60^{\circ} C$ to $30^{\circ} C$ in time $=t'$
Hence, we get:
$\frac{2.3026}{K} \log _{10} \frac{60-20}{30-20}=-t$
$\frac{-2.3026}{t} \log _{10} 4=K\dots (iii)$
Equating equations ( $i i$ ) and (iii), we get:
$\frac{-2.3026}{t} \log _{10} 4=\frac{-2.3026}{300} \log _{10} 2$
$t=300 \times 2=600 s =10 min$
Therefore, the time taken to cool the body from $60\,^{\circ} C$ to $30\,^{\circ} C$ is $10$ minutes.
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