Two identical beakers A and B contain equal volumes of two different liquids at $60^\circ\text{C}$ each and left to cool down. Liquid in A has density of $8\times10^2\ \text{kg m}^{-3}$ and specific heat of $2000\ \text{J kg}^{-1}\text{K}^{-1}$ while the liquid in B has density of $10^3\ \text{kg m}^{-3}$ and specific heat of $4000\ \text{J kg}^{-1}\text{K}^{-1}$. Which of the following best describes their temperature versus time graph schematically? (assume the emissivity of both the beakers to be the same)
Answer: (B) see figure
Both beakers lose heat at the same rate at the same temperature (identical beakers, same emissivity). The rate of fall of temperature is
$$\frac{dT}{dt} = -\frac{(\text{heat loss rate})}{\rho Vc}$$
Heat capacity per unit volume: A: $\rho c = 800\times2000 = 1.6\times10^6$; B: $1000\times4000 = 4\times10^6\ \text{J m}^{-3}\text{K}^{-1}$.
A has the smaller heat capacity, so it cools faster: at every later time A is colder than B, and the curves do not cross. The graph with B above A is correct.
Solution by Sreeraj P, M.Sc Physics