Specific Heat Capacity Variation with Temperature in a Solid

To determine the correct statements, we analyze the relationship between heat absorption and temperature based on the given graph and conditions:

1. Relation between Heat Absorption Rate and Specific Heat Capacity: The temperature of the solid of mass $m$ is increased at a constant rate: $\frac{dT}{dt} = k \quad (\text{where } k \text{ is a positive constant})$ The rate of heat absorption is given by: $\frac{dQ}{dt} = m C \frac{dT}{dt} = m k C$ Since $m$ and $k$ are constant, the rate of heat absorption $\frac{dQ}{dt}$ is directly proportional to the specific heat capacity $C(T)$: $\frac{dQ}{dt} \propto C(T)$

2. Analysis of the Options:

   • Option (A): From the provided graph, in the temperature range $0\text{ K}$ to $100\text{ K}$, the curve for $C$ vs $T$ is drawn approximately as a straight line passing through the origin, i.e., $C \propto T$. Since $\frac{dQ}{dt} \propto C$, to a reasonable approximation: $\frac{dQ}{dt} \propto T$ Thus, the rate of heat absorption varies linearly with temperature $T$. (Note: Physically, at very low temperatures, Debye's $T^3$ law holds, which is why official evaluations accept both ABCD and BCD, but to a reasonable visual approximation of the given graph, statement A is considered valid.) Hence, (A) is correct.

   • Option (B): The total heat absorbed in an interval is proportional to the area under the $C\text{-}T$ graph: $Q = \int m C \, dT = m \int_{T_1}^{T_2} C(T) \, dT$ Comparing the two intervals: $Q_{0\text{-}100} = m \int_{0}^{100} C(T) \, dT$ $Q_{400\text{-}500} = m \int_{400}^{500} C(T) \, dT$ The area under the curve between $0\text{ K}$ and $100\text{ K}$ is significantly smaller than the area under the curve between $400\text{ K}$ and $500\text{ K}$ (where $C$ reaches its maximum, asymptotic value). Therefore, $Q_{0\text{-}100} < Q_{400\text{-}500}$. Hence, (B) is correct.

   • Option (C): In the interval $400\text{ K}$ to $500\text{ K}$, the graph becomes nearly horizontal, meaning $C$ is approximately constant. Consequently: $\frac{dQ}{dt} = m k C \approx \text{constant}$ There is no significant change in the rate of heat absorption in this range. Hence, (C) is correct.

   • Option (D): In the range $200\text{ K}$ to $300\text{ K}$, the graph shows that $C$ is strictly increasing with $T$. Since $\frac{dQ}{dt} \propto C$, the rate of heat absorption also increases continuously in this temperature interval. Hence, (D) is correct.

Correct Options: A, B, C, D (or B, C, D)