$\frac{\mathrm{d}\theta}{\mathrm{d}t} = \frac{\sigma A}{mc}(\mathrm{T}^4 - \mathrm{T}_0^4).$ If the liquids put in exactly similar calorimeters and identical surrounding then we can consider $\mathrm{T}_0$ and A constant then $\frac{\mathrm{d}\theta}{\mathrm{d}t} \propto \frac{(\mathrm{T}^4 - \mathrm{T}_0^4)}{mc} \ldots \ldots (i)$ If we consider that equal masses of liquid (m) are taken at the same temperature then $\frac{\mathrm{d}\theta}{\mathrm{d}t} \propto \frac{1}{c}$ So, for same rate of cooling c should be equal which is not possible because liquids are of different nature. Again from equation (i) $\frac{\mathrm{dT}}{\mathrm{d}t} \propto \frac{(\mathrm{T}^4 - \mathrm{T}_0^4)}{mc} \implies \frac{\mathrm{d}\theta}{\mathrm{d}t} \propto \frac{(\mathrm{T}^4 - \mathrm{T}_0^4)}{V \rho c}$ Now if we consider that equal volume of liquid (V) are taken at the same temperature then $\frac{\mathrm{dT}}{\mathrm{d}t} \propto \frac{1}{\rho c}.$ So, for same rate of cooling multiplication of $\rho \times c$ for two liquid of different nature can be possible.