A liquid cools down from 70^{circ}C to 60^{ci | vedclass

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(C) According to Newton's law of cooling,the rate of cooling is directly proportional to the temperature difference between the body and its surroundi

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(C) According to Newton's law of cooling,the rate of cooling is directly proportional to the temperature difference between the body and its surroundings: $\frac{d	heta}{dt} = k(	heta - 	heta_0)$.In the first case,the liquid cools from $70^{\circ}C$ to $60^{\circ}C$. The average temperature is $\frac{70+60}{2} = 65^{\circ}C$. The temperature difference is $(65 - 	heta_0)$.In the second case,the liquid cools from $60^{\circ}C$ to $50^{\circ}C$. The average temperature is $\frac{60+50}{2} = 55^{\circ}C$. The temperature difference is $(55 - 	heta_0)$.Since the average temperature in the second case is lower,the temperature difference between the liquid and the surroundings is smaller. Consequently,the rate of cooling decreases. Therefore,it will take more time to cool through the same temperature range of $10^{\circ}C$.

$A$ liquid cools down from $70^{\circ}C$ to $60^{\circ}C$ in $5$ minutes. The time taken to cool it from $60^{\circ}C$ to $50^{\circ}C$ will be

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