A thermally conducting, uniform and homogeneous bar of length L, cross section A, density ρand

step 1 Okay, we have to calculate the reduced temperatures just like the previous ones. Similarly, redu

A thermally conducting, uniform and homogeneous bar of...

A thermally conducting, uniform and homogeneous bar of length $L_{\text {, }}$ cross section $A$, density $\rho$ and specific heat at constant pressure $c_p$ is brought to a nonuniform temperature distribution by contact at one end with a hot reservoir at a temperature $T_{\mathrm{H}}$ and at the other end with a cold reservoir at a temperature $T_{\mathrm{c}}$. The bar is removed from the reservoirs, thermally insulated and kept at constant pressure. Show that the change in entropy of the bar is $$ \Delta S=C_p\left(1+\ln T_{\mathrm{f}}+\frac{T_{\mathrm{c}}}{T_{\mathrm{H}}-T_{\mathrm{c}}} \ln T_{\mathrm{c}}-\frac{T_{\mathrm{H}}}{T_{\mathrm{H}}-T_{\mathrm{c}}} \ln T_{\mathrm{H}}\right), $$ where $C_p=c_p \rho A L, \quad T_{\mathrm{f}}=\left(T_{\mathrm{H}}+T_{\mathrm{c}}\right) / 2$

A thermally conducting, uniform and homogeneous bar of length $L_{\text {, }}$ cross section $A$, density $\rho$ and specific heat at constant pressure $c_p$ is brought to a nonuniform temperature distribution by contact at one end with a hot reservoir at a temperature $T_{\mathrm{H}}$ and at the other end with a cold reservoir at a temperature $T_{\mathrm{c}}$. The bar is removed from the reservoirs, thermally insulated and kept at constant pressure. Show that the change in entropy of the bar is
$$
\Delta S=C_p\left(1+\ln T_{\mathrm{f}}+\frac{T_{\mathrm{c}}}{T_{\mathrm{H}}-T_{\mathrm{c}}} \ln T_{\mathrm{c}}-\frac{T_{\mathrm{H}}}{T_{\mathrm{H}}-T_{\mathrm{c}}} \ln T_{\mathrm{H}}\right),
$$
where $C_p=c_p \rho A L, \quad T_{\mathrm{f}}=\left(T_{\mathrm{H}}+T_{\mathrm{c}}\right) / 2$

Key Concepts

Integration of dQ/T and Temperature-Dependent Entropy Change

The differential form dS = dQ/T is fundamental to calculating entropy change. Under constant pressure, where dQ can be expressed in terms of C? dT, integrating dS over the temperature range provides a method for quantifying entropy changes in a system undergoing nonisothermal conditions until equilibrium is reached.

Thermal Equilibration of Isolated Systems

When a system with a nonuniform temperature distribution is isolated, it naturally evolves towards a uniform temperature distribution, reaching thermal equilibrium. The redistribution of energy within the system, despite being irreversible, results in an overall change in entropy that can be calculated through state functions and reversible paths.

Entropy as a State Function

Because entropy is a state function, its change between an initial and final state is independent of the actual irreversible path taken by the system. This allows us to compute the entropy difference using a hypothetical reversible process, even though the actual process of reaching thermal equilibrium in the bar is irreversible.

Specific Heat at Constant Pressure

Specific heat at constant pressure, denoted as c?, quantifies how much energy is needed to raise the temperature of a unit mass by one degree while maintaining constant pressure. In this context, the total heat capacity C? = c? ? A L represents the energy capacity of the entire bar and is used in the integration to determine the change in entropy as the temperature changes.

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