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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 that temperature $T$ versus time $t$ plot for the two materials. We denote the specific heat of materials in the liquid (solid) states to be $C_{L 1}$ $\left(C_{S 1}\right)$ and $C_{L 2}\left(C_{S 2}\right)$, respectively. Choose the correct option given below.
$C_{L 1} < C_{L 2}$ and $C_{S 1} < C_{S 2}$
$C_{L 1} > C_{L 2}$ and $C_{S 1} < C_{S 2}$
$C_{L 1} > C_{L 2}$ and $C_{S 1} > C_{S 2}$
$C_{L 1} < C_{L 2}$ and $C_{S 1} > C_{S 2}$
Solution
(b)
Let $Q=$ rate of heat removal.
Then, $Q \cdot t=m c T$
$\Rightarrow T=\frac{Q}{m c} \cdot t$
Comparing this with $y=m x$,
Slope of $T-t$ line $\propto \frac{1}{\text { Specific heat }}$
From graph,
$\therefore C_{S 1} < C_{S 2}$
$\text { and } C_{L 1} > C_{L 2}$
