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

Options

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
D. Equal volumes of the liquids at the same temperature are taken

Answer

D

Explanation

`(dT)/(dt)=(sigmaA)/(mc)(T^4-T_0^4)`. If the liquids are put in exactly similar calorimeters and identical surrounding then we can cosider `T_0` and A constant. Then
`(dT)/(dt)prop((T^4-T_0^4))/(mc)` ..(i)
If we consider that equal masses of liquids (m) are taken at the same temperature then
`(dT)/(dt)prop((T^4-T_0^4))/(mc)`
`implies(dT)/(dt)prop((T^4-T_0^4))/(Vrhoc)`
Now if we consider that equal volumes of liquids `(V)` are taken at the same temperature then
`(dT)/(dt)prop(1)/(rhoc)`
so for same rate of cooling multiplication of `rhoxxc` for two liquids of different nature can be possible. so option (d) may be correct.