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(B) Let $P$ be the rate of heat removal (power). Since heat is extracted at the same rate,$P$ is constant for both.The heat lost by a substance is giv

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$C_{L1} > C_{L2}$ and $C_{S1} < C_{S2}$

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(B) Let $P$ be the rate of heat removal (power). Since heat is extracted at the same rate,$P$ is constant for both.The heat lost by a substance is given by $dQ = m \cdot c \cdot dT$,where $m$ is the mass and $c$ is the specific heat.The rate of heat removal is $P = \frac{|dQ|}{dt} = m \cdot c \cdot \left| \frac{dT}{dt} ight|$.Thus,the magnitude of the slope of the $T-t$ graph is $\left| \frac{dT}{dt} ight| = \frac{P}{m \cdot c}$.Since $P$ and $m$ are constant,the slope is inversely proportional to the specific heat: $	ext{Slope} \propto \frac{1}{c}$.$1$. For the liquid state (the initial cooling phase): Looking at the graph,the slope of curve $1$ is smaller than the slope of curve $2$. Since $	ext{Slope} \propto \frac{1}{C_L}$,a smaller slope implies a larger specific heat. Therefore,$C_{L1} > C_{L2}$.$2$. For the solid state (the final cooling phase): Looking at the graph,the slope of curve $1$ is larger than the slope of curve $2$. Since $	ext{Slope} \propto \frac{1}{C_S}$,a larger slope implies a smaller specific heat. Therefore,$C_{S1} Thus,the correct option is $C_{L1} > C_{L2}$ and $C_{S1} < C_{S2}$.

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