A container contains hot water at 100^{circ}C | 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

Vedclass · Accepted Answer

$T_1 < T_2$

Vedclass · Answer

(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.This implies that the time taken to cool down depends on the temperature difference between the body and the surroundings.For the first interval,the average temperature is $\frac{100+80}{2} = 90^{\circ}C$.For the second interval,the average temperature is $\frac{80+60}{2} = 70^{\circ}C$.Since the temperature difference between the water and the surroundings is greater in the first interval,the rate of cooling is faster.Therefore,the time taken to cool down by $20^{\circ}C$ is less in the first interval than in the second interval.Thus,$T_1 < T_2$.

$A$ container contains hot water at $100^{\circ}C$. If in time $T_1$ the temperature falls to $80^{\circ}C$ and in time $T_2$ the temperature falls to $60^{\circ}C$ from $80^{\circ}C$,then:

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