The rates of cooling of two different liquids | vedclass

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(D) According to Newton's law of cooling,the rate of cooling is given by $\frac{d	heta}{dt} = \frac{\sigma A}{mc}(T^4 - T_0^4)$.Since the calorimeter

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Equal volumes of the liquids at the same temperature are taken

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(D) According to Newton's law of cooling,the rate of cooling is given by $\frac{d	heta}{dt} = \frac{\sigma A}{mc}(T^4 - T_0^4)$.Since the calorimeters and surroundings are identical,$A$ and $T_0$ are constant. Thus,$\frac{d	heta}{dt} \propto \frac{1}{mc}$.Substituting mass $m = Vho$,where $V$ is volume and $ho$ is density,we get $\frac{d	heta}{dt} \propto \frac{1}{Vho c}$.For the rate of cooling to be the same for two different liquids,the product of density and specific heat capacity $(ho c)$ must be equal,or the volumes must be adjusted such that $V_1 ho_1 c_1 = V_2 ho_2 c_2$. Given the options,if we take equal volumes $(V_1 = V_2)$,the condition depends on the thermal properties. In many standard physics problems of this type,taking equal volumes is the intended condition to compare the cooling behavior based on their thermal capacities per unit volume. Thus,option $(d)$ is the correct choice.

The rates of cooling of two different liquids put in exactly similar calorimeters and kept in identical surroundings are the same if

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