In the irreversible process of Fig. 20.5, let the initial temperatures of the identical blocks L and R be 305.5Kand 294.5K, respectively, and let 215Jbe the energy that must be transferred between the blocks in order to reach equilibrium. For the reversible processes of Fig. 20.6,(a)What is $\mathbf{∆}\mathit{S}$for block L(b)What is $\mathbf{∆}\mathit{S}$for its reservoir, (c)What is $\mathbf{∆}\mathit{S}$for block R ,(d)What is$\mathbf{∆}\mathit{S}$for its reservoir, (e)What is$\mathbf{∆}\mathit{S}$for the two-block system, and (f)What is$\mathbf{∆}\mathit{S}$for the system of the two blocks and the two reservoirs?

1. The initial temperature of identical block, L,$T_{iL}=305.5\mathrm{K}$
2. The initial temperature of identical block, R,$T_{iR}=294.5\mathrm{K}$
3. The energy transferred between the identical blocks,$Q=215\mathrm{J}$

Entropy change is a phenomenon that quantifies how disorder or randomness has changed in a thermodynamic system. We can write the final temperature of the system from the initial temperatures of the two blocks. Then using the formula for heat added, we can find the masses of the blocks. By using the formula for entropy change, we can find the change in entropy of blocks L and R, their corresponding reservoirs, the system of blocks, and the entire system.

Formulae:

The heat transferred by the body,$Q=mc∆T$ …(i)

The entropy change of the gas, $∆S=mcIn\left(\frac{T_f}{T_i}\right)$ …(ii)

At equilibrium, since the two blocks are identical, their final temperature is given as:

$$\begin{gathered} T_f=\frac{T_{iL}+T_{iR}}{2} \\ =\frac{305.5\mathrm{K}+294.5\mathrm{K}}{2} \\ =300.0\mathrm{K} \end{gathered}$$

Substituting the given values in equation (i), we can get the mass of the block as given:

$$\begin{gathered} m=\frac{Q}{c∆T} \\ =\frac{215\mathrm{J}}{386\mathrm{J}/\mathrm{kgK}\left(300\mathrm{K}-294.5\mathrm{K}\right)} \\ =0.10127\mathrm{kg} \\ ~0.101\mathrm{kg} \end{gathered}$$

Change in the entropy of block L using equation (ii) is given as:

$$\begin{gathered} ∆S_L=mcIn\left(\frac{T_f}{T_{iL}}\right) \\ =\left(0.10127\mathrm{kg}\right)\left(386\mathrm{J}/\mathrm{kgK}\right)\mathrm{In}\left(\frac{300.0\mathrm{K}}{305.5\mathrm{K}}\right) \\ =-0.710\mathrm{J}/\mathrm{K} \end{gathered}$$

Hence, the value of the entropy change in block, L is$-0.710\mathrm{J}/\mathrm{K}$

Change in entropy of the reservoir of block L = - Change in entropy of block L

$∆S_l=0.710\frac{\mathrm{J}}{\mathrm{K}}$

Hence, the value of the entropy change is$0.710\frac{\mathrm{J}}{\mathrm{K}}$

Substituting the given values in equation (ii), we can get the change in the entropy of the block, R is given as follows:

$$\begin{gathered} ∆S_R=\left(0.10127\mathrm{kg}\right)\left(386\mathrm{J}/\mathrm{kgK}\right)\mathrm{In}\left(\frac{300.0\mathrm{K}}{294.5\mathrm{K}}\right) \\ =0.723\mathrm{J}/\mathrm{K} \end{gathered}$$

Hence, the value of the entropy change of block, R is$0.723\mathrm{J}/\mathrm{K}$

Change in entropy of the reservoir of block R = - Change in entropy of block R

$∆S_r=-0.723\frac{\mathrm{J}}{\mathrm{K}}$

Hence, the value of the change in entropy is$-0.723\frac{\mathrm{J}}{\mathrm{K}}$

Change in entropy of the two block system = Change in entropy of block L + Change in entropy of block R

$$\begin{gathered} ∆S=∆S_L+∆S_R \\ =-0.710\mathrm{J}/\mathrm{K}+0.723\mathrm{J}/\mathrm{K} \\ =0.013\mathrm{J}/\mathrm{K} \end{gathered}$$

Hence, the value of the entropy change of the two-block system is$0.013\mathrm{J}/\mathrm{K}$

Change in entropy of the system of two blocks and the two reservoirs is given as:

$$\begin{gathered} ∆S=∆S_L+∆S_R+∆S_{lr}+∆S_{rr} \\ =-0.710\mathrm{J}/\mathrm{K}+0.723\mathrm{J}/\mathrm{K}+0.710\mathrm{J}/\mathrm{K}-0.723\mathrm{J}/\mathrm{K} \\ =0 \end{gathered}$$

Hence, the value of the entropy change of the system is zero.