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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 below shows the temperature $T$ versus time $t$ plot for the two materials. We denote the specific heat in the liquid states to be $C_{L 1}$ and $C_{L 2}$ for materials 1 and 2, respectively and latent heats of fusion $U_1$ and $U_2$, respectively. Choose the correct option.
$C_{L 1} > C_{L 2}$ and $U_1 < U_2$
$C_{L 1} > C_{L 2}$ and $U_1 > U_2$
$C_{L 1} < C_{L 2}$ and $U_1 > U_2$
$C_{L 1} < C_{L 2}$ and $U_1 < U_2$
Solution
(c)
Wehave, heat extracted from a liquid during solidification,
$U=Q t=m L \Rightarrow L \propto U$
Also, heat extracted from liquid during cooling,
$H=Q t=m c \Delta T$
Temperature of liquid,
$T=\frac{Q}{m c} \cdot t+T_i$
Slope of $T$ versus $t$ line is inversely proportional to specific heat $c$.
Now, from given graph, we get
we get, $U_1 > U_2$ and $C_{L 1} < C_{L 2}$
