91Ó°ÊÓ

The given amount of energy conducted by the reservoir, Q = 260 J

The constant temperature, ${\mathrm{T}}_1=400\mathrm{K}$.

The temperatures,${\mathrm{T}}_2=100\mathrm{K},{\mathrm{T}}_2=200\mathrm{K},{\mathrm{T}}_3=300\mathrm{K}\mathrm{and}{\mathrm{T}}_4=360\mathrm{K}$.

By using the formula for the change in the entropy, we can find the net change in entropy of reservoirs in each of the given cases; also from this, if the temperature difference between the two reservoirs decreases, we can find if the net change in entropy increases, decreases, or remains the same.

Formula:

The change in the entropy for a given amount of energy of the cycle,

$∆S=Q\left(\frac{1}{{\mathrm{T}}_2}-\frac{1}{{\mathrm{T}}_1}\right)$ (1)

The change in the entropy is transferring a certain amount of heat Q from a heat reservoir at${\mathrm{T}}_1$to another one at${\mathrm{T}}_2=100\mathrm{K}$is given using equation (1) as follows:

$$\begin{gathered} ∆S=260\mathrm{J}×\left(\frac{1}{100\mathrm{K}}-\frac{1}{400\mathrm{K}}\right) \\ =1.95\mathrm{J}/\mathrm{K} \end{gathered}$$

Hence, the value of the entropy change is 1.95 J/K

The change in the entropy is transferring a certain amount of heat Q from a heat reservoir at${\mathrm{T}}_1$to another one at${\mathrm{T}}_2=200\mathrm{K}$is given using equation (1) as follows:

$$\begin{gathered} ∆\mathrm{S}=260\mathrm{J}×\left(\frac{1}{200\mathrm{K}}-\frac{1}{400\mathrm{K}}\right) \\ =0.650\mathrm{J}/\mathrm{K} \end{gathered}$$

Hence, the value of the entropy change is 0.650 J/K

The change in the entropy is transferring a certain amount of heat Q from a heat reservoir at${\mathrm{T}}_1$to another one at${\mathrm{T}}_3=300\mathrm{K}$is given using equation (1) as follows:

$$\begin{gathered} ∆\mathrm{S}=260\mathrm{J}×\left(\frac{1}{300\mathrm{K}}-\frac{1}{400\mathrm{K}}\right) \\ =0.217\mathrm{J}/\mathrm{K} \end{gathered}$$

Hence, the value of the entropy change is 0.217 J/K

The change in the entropy is transferring a certain amount of heat Q from a heat reservoir at ${\mathrm{T}}_1$to another one at ${\mathrm{T}}_4=360\mathrm{K}$is given using equation (1) as follows:

$$\begin{gathered} ∆\mathrm{S}=260\mathrm{J}×\left(\frac{1}{360\mathrm{K}}-\frac{1}{400\mathrm{K}}\right) \\ =0.072\mathrm{J}/\mathrm{K} \end{gathered}$$

Hence, the value of the entropy change is 0.072 J/K

We see from the calculations of the above parts that as the temperature difference between the two reservoirs decreases, the net change in the entropy $∆{\mathrm{S}}_{\mathrm{net}}$also decreases.