Question

Short answer

(a) \(68\% \) (b) \(0.43\;{\rm{J/K}}\) and (c) \(0\)

Step-by-step solution

Given Data

The lower temperature of reservoir is\({T_{\rm{L}}} = 650\;{\rm{K}}\).

The higher temperature of reservoir is\({T_{\rm{H}}} = 970\;{\rm{K}}\).

The work is\(W = 550\;{\rm{J}}\).

The heat input is \(Q = 2500\;{\rm{J}}\).

Understanding the efficiency of real engine

In this problem, the efficiency of the real engine can be evaluated using the fraction between the work done by the engine and heat input.

Comparison between the real engine and Carnot engine efficiency  

The relation of real engine efficiency is

\(\begin{array}{l}\eta = \frac{W}{Q}\\\eta = \left( {\frac{{550\;{\rm{J}}}}{{2500\;{\rm{J}}}}} \right)\\\eta = 0.22.\end{array}\)

The relation of ideal engine efficiency is

\(\begin{array}{l}{\eta _{\rm{C}}} = 1 - \frac{{{T_{\rm{L}}}}}{{{T_{\rm{H}}}}}\\{\eta _{\rm{C}}} = 1 - \left( {\frac{{650\;{\rm{K}}}}{{970\;{\rm{K}}}}} \right)\\{\eta _{\rm{C}}} = 0.32.\end{array}\)

The comparison between efficiency is

\(\begin{array}{l}\frac{\eta }{{{\eta _{\rm{C}}}}} = \frac{{0.22}}{{0.32}}\\\frac{\eta }{{{\eta _{\rm{C}}}}} = 0.68 \times 100\% \\\frac{\eta }{{{\eta _{\rm{C}}}}} = 68\% .\end{array}\)

Thus, the actual efficiency is \(68\% \) of ideal efficiency.

Calculation of entropy change in the real engine

The relation of total entropy change is given below:

\(\begin{array}{l}\Delta S = \Delta {S_{\rm{i}}} + \Delta {S_{\rm{0}}}\\\Delta S = - \left( {\frac{Q}{{{T_{\rm{H}}}}}} \right) + \left( {\frac{{Q'}}{{{T_{\rm{L}}}}}} \right)\\\Delta S = - \left( {\frac{Q}{{{T_{\rm{H}}}}}} \right) + \left( {\frac{{Q - W}}{{{T_{\rm{L}}}}}} \right)\end{array}\)

Here,\(\Delta {S_{\rm{i}}}\)and\(\Delta {S_{\rm{0}}}\)are the entropies at input and output, and\(Q'\)is the heat loss from the reservoir.

Put the values in the above relation.

\(\begin{array}{l}\Delta S = - \left( {\frac{{2500\;{\rm{J}}}}{{970\;{\rm{K}}}}} \right) + \left( {\frac{{2500\;{\rm{J}} - 550\;{\rm{J}}}}{{650\;{\rm{K}}}}} \right)\\\Delta S = 0.43\;{\rm{J/K}}\end{array}\)

Thus, \(\Delta S = 0.43\;{\rm{J/K}}\) is the required change in entropy.

Calculation of entropy change in the ideal engine

The relation of total entropy change is shown below:

\(\begin{array}{l}\Delta S = \Delta {S_{\rm{i}}} + \Delta {S_{\rm{0}}}\\\Delta S = - \left( {\frac{Q}{{{T_{\rm{H}}}}}} \right) + \left( {\frac{{Q'}}{{{T_{\rm{L}}}}}} \right)\end{array}\)

Here,\(Q'\)is the heat loss from the reservoir in the ideal engine whose value is\(Q' = \left( {\frac{{Q{T_{\rm{L}}}}}{{{T_{\rm{H}}}}}} \right)\).

Put the values in the above relation.

\(\begin{array}{l}\Delta S = - \left( {\frac{Q}{{{T_{\rm{H}}}}}} \right) + \left( {\frac{{\left( {\frac{{Q{T_{\rm{L}}}}}{{{T_{\rm{H}}}}}} \right)}}{{{T_{\rm{L}}}}}} \right)\\\Delta S = - \left( {\frac{Q}{{{T_{\rm{H}}}}}} \right) + \left( {\frac{Q}{{{T_{\rm{H}}}}}} \right)\\\Delta S = 0\end{array}\)

Thus, \(\Delta S = 0\) is the required change in entropy.