A brass rod is in thermal contact with a constant-temperature reservoir at$\mathbf{130}\mathbf{°}\mathbf{C}$at one end and a constant-temperature reservoir at $\mathbf{24}\mathbf{.}\mathbf{0}\mathbf{°}\mathit{C}$ at the other end. (a) Compute the total change in entropy of the rod–reservoirs system when 5030 Jof energy is conducted through the rod, from one reservoir to the other. (b) Does the entropy of the rod change?

The temperature at one end,${\mathrm{T}}_{\mathrm{cold}}=130°\mathrm{C}=403.15\mathrm{K}$

The temperature at the other end,${\mathrm{T}}_{\mathrm{hot}}=24°\mathrm{C}=297.15\mathrm{K}$

Energy conducted through the rod, $\mathrm{Q}=5030\mathrm{J}$

By using the concept of change in entropy from the equation, we can find the entropy change for each reservoir. This will help in determining the net entropy change of the rod.

Formula:

The entropy change of a system from the second law of thermodynamics,

$∆\mathrm{S}=\frac{\mathrm{Q}}{\mathrm{T}}$ (1)

The net entropy change of the rod-reservoir system can be given using the given data in equation (1) as follows: ($∆{\mathrm{S}}_{\mathrm{cold}}$ is negative because heat is removed from the substance.)

role="math" localid="1661337705241" $$\begin{gathered} ∆\mathrm{S}=∆{\mathrm{S}}_{\mathrm{hot}}+∆{\mathrm{S}}_{\mathrm{cold}} \\ =\frac{\mathrm{Q}}{{\mathrm{T}}_{\mathrm{hot}}}-\frac{\mathrm{Q}}{{\mathrm{T}}_{\mathrm{cold}}} \\ =\frac{5030\mathrm{J}}{297.15\mathrm{K}}-\frac{5030\mathrm{J}}{403.15\mathrm{K}} \\ =\frac{5030\mathrm{J}\times 403.15\mathrm{K}-5030\mathrm{J}\times 297.15\mathrm{K}}{297.15\mathrm{K}\times 403.15\mathrm{K}} \\ =\frac{5030\mathrm{J}\times 106\mathrm{K}}{12\times {10}^4{\mathrm{K}}^2} \\ =4.45\mathrm{J}/\mathrm{K} \end{gathered}$$

Hence, the value of the total entropy change of the system is 4.45 J/K

We assumed that the heat flow in the rod is a steady state. If the process is reversible, then there is no change in the entropy. It is given that the process is reversible, so there is no change in the entropy.Therefore, no entropy change occurs in the rod for the thermodynamic process,

i.e.$∆{\mathrm{S}}_{\mathrm{rod}}=0$

Hence, the value of the entropy of the rod is zero.