The ratio of the heat capacities of two bodie | vedclass

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(B) The rate of heat loss is given by $P = \frac{dQ}{dt}$.We know that $dQ = C \cdot dT$,where $C$ is the heat capacity and $dT$ is the change in temp

Vedclass · Accepted Answer

$4 : 1$

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(B) The rate of heat loss is given by $P = \frac{dQ}{dt}$.We know that $dQ = C \cdot dT$,where $C$ is the heat capacity and $dT$ is the change in temperature.Therefore,the rate of heat loss is $P = C \cdot \frac{dT}{dt}$.Given that the rate of heat loss $P$ is the same for both bodies,we have $C_1 \cdot \left(\frac{dT}{dt}\right)_1 = C_2 \cdot \left(\frac{dT}{dt}\right)_2$.This implies $\frac{(\frac{dT}{dt})_1}{(\frac{dT}{dt})_2} = \frac{C_2}{C_1}$.Given the ratio of heat capacities $\frac{C_1}{C_2} = \frac{1}{4}$,we find $\frac{C_2}{C_1} = \frac{4}{1}$.Thus,the ratio of the rate of fall of temperatures is $4:1$.

The ratio of the heat capacities of two bodies is $1:4$. If the rate of heat loss is the same for both bodies under similar environmental conditions,then the ratio of the rate of fall of their temperatures is:

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