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(C) The rate of heat extraction is constant, let it be $P$. For the liquid cooling phase, the heat released is $H = m C_L \Delta T$. Since $H = P \cdo

Vedclass · Accepted Answer

$C_{L 1}  L_2$

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(C) The rate of heat extraction is constant, let it be $P$. For the liquid cooling phase, the heat released is $H = m C_L \Delta T$. Since $H = P \cdot t$, we have $P \cdot t = m C_L \Delta T$, which gives the slope of the $T-t$ graph as $\frac{\Delta T}{t} = \frac{P}{m C_L}$. Since $P$ and $m$ are the same for both, the slope is inversely proportional to the specific heat $C_L$. From the graph, the slope of line $2$ is steeper than that of line $1$, so $C_{L 2} > C_{L 1}$, or $C_{L 1} For the phase change (solidification) part, the heat released is $H = m L$. Since $H = P \cdot t_{phase}$, we have $P \cdot t_{phase} = m L$, which gives $L = \frac{P}{m} t_{phase}$. Thus, the latent heat $L$ is directly proportional to the time $t_{phase}$ spent in the phase change. From the graph, the horizontal segment for material $1$ is longer than that for material $2$, so $t_{phase, 1} > t_{phase, 2}$, which implies $L_1 > L_2$. Therefore, $C_{L 1}  L_2$. The correct option is $C$.

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