Question

Suppose that the entropy of a certain substance (not anEinstein solid) is given by $\mathbf{S}\mathbf{=}\mathbf{a}\sqrt{\mathbf{E}}$, where ais a constant. Whatis the specific heat capacity Cas a function of the temperature T?

Short answer

The specific heat capacity as a function of temperature can be given as,$\mathrm{C}=\frac{{\mathrm{a}}^2\mathrm{T}}{2}$

Step-by-step solution

Identification of given data

The entropy of substance,$\mathrm{S}=\mathrm{a}\sqrt{\mathrm{E}}$

Where a is constant, E is energy and T is temperature

Relation is used to solve the problem.

• Definition of the temperature.

$\frac{\mathbf{1}}{\mathbf{T}}\mathbf{=}\frac{\mathbf{\partial }\mathbf{S}}{\mathbf{\partial }\mathbf{E}}$

Where T is temperature,$\mathbf{\partial }\mathbf{S}$ is the change in entropy, and$\mathbf{\partial }\mathbf{E}$ is the change in energy.

• Definition of specific heat capacity.

$\mathbf{C}\mathbf{=}\frac{\mathbf{\partial }\mathbf{E}}{\mathbf{\partial }\mathbf{T}}$

Where Cis specific heat capacity,$\mathbf{\partial }\mathbf{T}$ is the temperature change, and$\mathbf{\partial }\mathbf{E}$ is the change in energy.

Finding the specific heat capacity as a function of time

Given in the question,

$\mathrm{S}=\mathrm{a}\sqrt{\mathrm{E}}$

From the definition of temperature we know,

$\frac{1}{\mathrm{T}}=\frac{\partial S}{\partial E}$

Substituting the value of S

$$\begin{gathered} \frac{1}{\mathrm{T}}=\frac{\partial \left(\mathrm{a}\sqrt{\mathrm{E}}\right)}{\partial E} \\ \frac{1}{\mathrm{T}}=\frac{\partial \left(\mathrm{a}\sqrt{\mathrm{E}}\right)}{\partial E} \\ \frac{1}{\mathrm{T}}=\frac{\partial \left(\mathrm{a}\sqrt{\mathrm{E}}\right)}{\partial E} \\ \frac{1}{\mathrm{T}}=\frac{\partial \left(\mathrm{a}\sqrt{\mathrm{E}}\right)}{\partial E} \\ \frac{1}{\mathrm{T}}=\frac{\mathrm{a}\partial \left(\sqrt{\mathrm{E}}\right)}{\partial E} \\ \frac{1}{\mathrm{T}}=\mathrm{a}\left(\frac{1}{2}{\mathrm{E}}^{\left(1/2\right)-1}\right) \\ \frac{1}{\mathrm{T}}=\frac{\mathrm{a}}{2\sqrt{\mathrm{E}}} \end{gathered}$$

Further simplification will give,

$$\begin{gathered} \sqrt{\mathrm{E}}=\frac{\mathrm{aT}}{2} \\ \mathrm{E}=\frac{{\mathrm{a}}^2{\mathrm{T}}^2}{4} \end{gathered}$$

The energy in terms of temperature can be given as, $\mathrm{E}=\frac{{\mathrm{a}}^2{\mathrm{T}}^2}{4}$

Now from the definition of heat capacity, we know,

$\mathrm{C}=\frac{\partial \mathrm{E}}{\partial \mathrm{T}}$

Now substituting the value of E.

$$\begin{gathered} \mathrm{C}=\frac{\partial \left({\mathrm{a}}^2{\mathrm{T}}^2/4\right)}{\partial \mathrm{T}} \\ \mathrm{C}=\frac{{\mathrm{a}}^2}{4}\frac{\partial \left({\mathrm{T}}^2\right)}{\partial \mathrm{T}} \\ \mathrm{C}=\frac{{\mathrm{a}}^2}{4}\left(2\mathrm{T}\right) \\ \mathrm{C}=\frac{{\mathrm{a}}^2\mathrm{T}}{2} \end{gathered}$$

The specific heat capacity is, $\mathrm{C}=\frac{{\mathrm{a}}^2\mathrm{T}}{2}$.