Animated Solution for Physics - Properties of Solids and Liquids: The figure below shows the variation of specific heat capacity ($C$) of a solid as a function of temperature ($T$). The temperature is increased continuously from $0$ to $500\text{ K}$ at a constant rate. Ignoring any volume change, the following statement(s) is (are) correct to reasonable approximation.

When we heat a solid, its temperature doesn't just rise arbitrarily. The way a material absorbs heat is governed by its specific heat capacity, $C$. In classical physics, we often assume $C$ is a constant. However, as we dive into the quantum realm at low temperatures, the specific heat behaves in fascinating ways, as beautifully illustrated in this problem's graph.

The problem provides us with a graph of specific heat capacity $C$ versus temperature $T$. A crucial piece of information is hidden in the text: "The temperature is increased continuously... at a constant rate."

Mathematically, this means the derivative of temperature with respect to time is a constant:

To evaluate the options, we need to understand the rate of heat absorption, which we will call $R$. From the fundamental principles of calorimetry, the small amount of heat $dQ$ absorbed by a mass $m$ to raise its temperature by $dT$ is:

To find the rate, we divide both sides by the time differential $dt$:

Substituting our constant rate $k$ into the equation, we get:

Since the mass $m$ and the heating rate $k$ are constants, we arrive at a powerful conclusion:

The rate of heat absorption is directly proportional to the specific heat capacity. This means the graph of $R$ versus $T$ will have the exact same shape as the given $C$ versus $T$ graph!

The Low-Temperature Quantum Realm (0-100 K)

Let's evaluate Option A. Look at the graph in the range of $0$ to $100\text{ K}$. Is it a straight line? No, it is a curve that starts flat and bends upwards. In solid-state physics, this is explained by the Debye Model, which states that at low temperatures, the specific heat is proportional to the cube of the temperature ($C\propto T^3$).

Because $C$ is not linear, and $R\propto C$, the rate of heat absorption $R$ also does not vary linearly. Therefore, Option A is incorrect.

The Area Under the Curve

Now, let's look at Option B, which compares the total heat absorbed in two different temperature intervals. The total heat absorbed $\Delta Q$ is the integral of the heat rate over time, or equivalently, the integral of specific heat over temperature:

Geometrically, the integral $\int CdT$ represents the area under the $C-T$ curve.

If we visually compare the area under the curve from $0$ to $100\text{ K}$ with the area from $400$ to $500\text{ K}$, the difference is stark. The area from $0$ to $100\text{ K}$ is a tiny sliver, while the area from $400$ to $500\text{ K}$ is a massive rectangle.

Since ${\text{Area}}_{0-100}<{\text{Area}}_{400-500}$, it strictly follows that:

The heat absorbed in the first interval is indeed less. Option B is absolutely correct.

The Classical Limit (400-500 K)

Let's evaluate Option C. In the range of $400$ to $500\text{ K}$, the graph becomes completely horizontal. The specific heat $C$ has reached a constant maximum value. This phenomenon is described by the Dulong-Petit Law, which states that at high temperatures, the molar heat capacity of a solid approaches $3R$ (where $R$ is the universal gas constant).

Since $C$ is constant in this range, and $R\propto C$, the rate of heat absorption $R$ is also perfectly constant. There is no change in the rate. Option C is correct.

The Transition Zone (200-300 K)

Finally, let's check Option D. In the interval from $200$ to $300\text{ K}$, the curve is sloping upwards. This means the specific heat $C$ is actively increasing as the solid absorbs more thermal energy and its atomic lattice vibrations become more excited.

Because $C$ is increasing, the rate of heat absorption $R$ must also be increasing. Option D is correct.

By seamlessly connecting the mathematical definition of heating rates with the geometric interpretation of graphs, we have successfully decoded the thermodynamic behavior of the solid. The correct statements are (b), (c), and (d).