• question_answer Two solid bodies of equal mass m initially at $T=0{}^\circ C$ are heated at a uniform and same rate under identical conditions. The temperature of the first object with latent heat ${{L}_{1}}$ and specific heat capacity in solid state ${{C}_{1}}$ changes according to graph 1 on the diagram. The temperature of the second object with latent heat ${{L}_{2}}$ and specific heat capacity in solid state ${{C}_{2}}$ changes according to graph 2 on the diagram. Based on what is shown on the graph, the latent heats ${{L}_{1}}$ and ${{L}_{2}}$, and the specific heat capacities ${{C}_{1}}$ and ${{C}_{2}}$ in solid state obey which of the following relationships : A) ${{L}_{1}}>{{L}_{2}};\text{ }{{C}_{1}}<{{C}_{2}}$ B) ${{L}_{1}}<{{L}_{2}};\text{ }{{C}_{1}}<{{C}_{2}}$ C) ${{L}_{1}}>{{L}_{2}};\text{ }{{C}_{1}}>{{C}_{2}}$ D) ${{L}_{1}}<{{L}_{2}};\text{ }{{C}_{1}}>{{C}_{2}}$

 If heat is supplied at constant rate P, then $Q=P\Delta t$and as during  change of state $Q=mL$, so, $mL=P\Delta t$ i.e.,       $L=\left[ \frac{P}{m} \right]\Delta t=\frac{P}{m}$ (length of line AB) Hence, ${{L}_{1}}>{{L}_{2}}$ i.e., the ratio of latent heat of fusion of the two substances are in the ratio 3:4. In the portion OA the substance is in solid state and its temperature is changing.
 $\Delta Q=mC\Delta T$ and       $\Delta Q=P\Delta t$ So, $\frac{\Delta T}{\Delta t}=\frac{P}{mC}$     or         $slope=\frac{P}{mS}$ $=\left[ as\frac{\Delta T}{\Delta t}=slope \right]\,\,;$ Hence ${{C}_{1}}<{{C}_{2}}$
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