Question

Suppose that 260 Jis conducted from a constant-temperature reservoir at 400 kto one at (a) 100 K, (b) 200 K, (c) 300 K, and (d) 360 K. What is the net change in entropy$\mathbf{∆}{\mathit{S}}_{\mathbf{n}\mathbf{e}\mathbf{t}}$of the reservoirs in each case? (e) As the temperature difference of the two reservoirs decreases, does$\mathbf{∆}{\mathit{S}}_{\mathbf{n}\mathbf{e}\mathbf{t}}$increase, decrease, or remain the same?

Short answer

The net change in entropy $∆S_{\mathrm{net}}$of reservoirs for temperatures:

$$\begin{gathered} \mathrm{a})100\mathrm{K}\mathrm{is}1.95\mathrm{J}/\mathrm{K} \\ \mathrm{b})200\mathrm{K}\mathrm{is}0.650\mathrm{J}/\mathrm{K} \\ \mathrm{c})300\mathrm{K}\mathrm{is}0.217\mathrm{J}/\mathrm{K} \\ \mathrm{d})360\mathrm{is}0.072\mathrm{J}/\mathrm{K} \end{gathered}$$

e) If the temperature difference between the two reservoirs decreases, the net change in entropy $∆{\mathrm{S}}_{\mathrm{net}}$also decreases.

Step-by-step solution

The given data

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

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

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

Understanding the concept of entropy change

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)

a) Calculation of entropy change for temperature 100 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${\mathrm{T}}_2=100\mathrm{K}$is given using equation (1) as follows:

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

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

b) Calculation of entropy change for temperature 200K

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:

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

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

c) Calculation of entropy change for temperature 300K

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:

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

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

d) Calculation of entropy change for temperature 360K

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:

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

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

e) Checking the behavior of entropy change with temperature change

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.