Question

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

• A. The masses of the liquids are equal
• B. Equal masses of the liquids at the same temperature are taken
• C. Different volumes of the liquids at the same temperature are taken
• D. Equal volumes of the liquids at the same temperature are taken

Answer 1

Answer: (D)

Equal volumes of the liquids at the same temperature are taken

Answer 2

$\frac{dT}{dt} = \frac{\sigma A}{mc}(T^{4} - T_{0}^{4})$. If the liquids put in exactly similar calorimeters and identical surrounding then we can consider T₀ and A constant then $\frac{dT}{dt} \propto \frac{(T^{4} - T_{0}^{4})}{mc}$……(i)

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

So for same rate of cooling c should be equal which is not possible because liquids are of different nature.

Again from (i) equation $\frac{dT}{dt} \propto \frac{(T^{4} - T_{0}^{4})}{mc}$ ⇒

$\frac{dT}{dt} \propto \frac{(T^{4} - T_{0}^{4})}{V\rho c}$

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 ρ× c for two liquid of different nature can be possible. So option (4) may be correct.