$\frac{\text{d}\theta }{\text{dt}}=\frac{\sigma \text{A}}{\text{mc}}\left({\text{T}}^4-{\text{T}}_0^4\right)$. If the liquids put in exactly similar calorimeters and identical surrounding then we can consider T₀ and A constant then $\frac{\text{d}\theta }{\text{dt}}\propto \frac{\left({\text{T}}^4-{\text{T}}_0^4\right)}{\text{mc}}$

If we consider that equal masses of liquid (m) are taken at the same temperature then $\frac{\text{d}\theta }{\text{dt}}\propto \frac{1}{\text{c}}$

So for same rate of cooling c should be equal which is not possible because liquids are of different nature. Again from equation (i)

$\frac{\text{d}\theta }{\text{dt}}\propto \frac{\left({\text{T}}^4-{\text{T}}_0^4\right)}{\text{mc}}\Rightarrow \frac{\text{d}\theta }{\text{dt}}\propto \frac{\left({\text{T}}^4-{\text{T}}_0^4\right)}{\text{V}\rho \text{c}}$

Now if we consider that equal volume of liquid (V) are taken at the same temperature then $\frac{\text{d}\theta }{\text{dt}}\propto \frac{1}{\rho \text{c}}$

So for same rate of cooling multiplication of $\rho \times \text{c}$ for two liquid of different nature can be possible