A body cools down from 75^{circ}C to 60^{circ | vedclass

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(C) According to Newton's law of cooling,the rate of cooling is given by $\frac{dT}{dt} = -k(T - T_s)$,where $T$ is the temperature of the body and $T

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(C) According to Newton's law of cooling,the rate of cooling is given by $\frac{dT}{dt} = -k(T - T_s)$,where $T$ is the temperature of the body and $T_s$ is the temperature of the surroundings.For the first interval,the average temperature is $T_{avg1} = \frac{75+60}{2} = 67.5^{\circ}C$.For the second interval,the average temperature is $T_{avg2} = \frac{65+55}{2} = 60^{\circ}C$.Since $T_{avg2} Therefore,the rate of cooling $\frac{dT}{dt}$ is lower in the second case.Since the temperature drop is $10^{\circ}C$ in both cases,a lower rate of cooling implies that it will take more time to cool down in the second case.

$A$ body cools down from $75^{\circ}C$ to $60^{\circ}C$ in $10 \text{ minutes}$. It will cool down from $65^{\circ}C$ to $55^{\circ}C$ in a time:

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