Question

Two different liquids of same mass are kept in two identical vessels, which are placed in a freezer that extracts heat from them at the same rate causing each liquid to transform into a solid. The schematic figure shows the temperature T vs time t plot for the two materials. We denote the specific heat in the liquid states to be $C_{L1}$ and $C_{L2}$ for materials $1$ and $2$ respectively, and latent heats of fusion $U_1$ and $U_2$ respectively. Choose the correct option.

• A. $C_{L1}>C_{L2}$ and $U_1<U_2$
• B. $C_{L1}>C_{L2}$ and $U_1>U_2$
• C. $C_{L1}<C_{L2}$ and $U_1>U_2$
• D. $C_{L1}<C_{L2}$ and $U_1<U_2$

Answer 1

The correct option is C $C_{L1}<C_{L2}$ and $U_1>U_2$

Rate of heat extracted $R=\frac{-m{C}_L\Delta T}{\Delta t}$
$⟹$ $\Delta T=\frac{-R}{m{C}_L}\Delta t$
Since, magnitude of slope of liquid 1 is greater than that of liquid 2.
So, $\frac{R}{m{C}_{L1}}>\frac{R}{m{C}_{L2}}$
where $m$ and $R$ are constants.
$⟹C_{L2}>C_{L1}$
Part of graph where $T$ is constant represents the phase change of substance.
Since, more time is required for liquid 1 to change into solid. So more heat is extracted from liquid 1 as compared to liquid 2.
where heat extracted $H=mU$
$⟹U_1>U_2$ $(\because H_1>H_2)$