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The rates of cooling of two different liquids put in exactly similar calorimeters and kept in identical surroundings are the same if
The masses of the liquids are equal
Equal masses of the liquids at the same temperature are taken
Different volumes of the liquids at the same temperature are taken
Equal volumes of the liquids at the same temperature are taken
Solution
(d) $\frac{{d\theta }}{{dt}} = \frac{{\sigma A}}{{mc}}({T^4} – T_0^4)$. If the liquids put in exactly similar calorimeters and identical surrounding then we can consider T0 and $A$ constant then $\frac{{d\theta }}{{dt}} \propto \frac{{({T^4} – T_0^4)}}{{mc}}$……$(i)$
If we consider that equal masses of liquid $(m)$ are taken at the same temperature then $\frac{{d\theta }}{{dt}} \propto \frac{1}{c}$
So for same rate of cooling c should be equal which is not possible because liquids are of different nature. Again from equation $(i)$
$\frac{{dT}}{{dt}} \propto \frac{{({T^4} – T_0^4)}}{{mc}}$ ==> $\frac{{d\theta }}{{dt}} \propto \frac{{({T^4} – T_0^4)}}{{V\rho \,c}}$
Now if we consider that equal volume of liquid $(V)$ are taken at the same temperature then $\frac{{dT}}{{dt}} \propto \frac{1}{{\rho \,c}}$.
So for same rate of cooling multiplication of $p× c$ for two liquid of different nature can be possible. So option $(d)$ may be correct.
