(a) In a two-stage Carnot heat engine, a quantity of heat \(\left|Q_{1}\right|\) is absorbed at a temperature \(T_{1}\), work \(\left|W_{1}\right|\) is done, and a quantity of heat \(\left|Q_{2}\right|\) is expelled at a lower temperature \(T_{2}\), by the first stage. The second stage absorbs the heat expelled by the first, does work \(\left|W_{2}\right|\), and expels a quantity of heat \(\left|Q_{3}\right|\) at a lower temperature \(T_{3}\). Prove that the efficiency of the combination is \(\left(T_{1}-T_{3}\right) / T_{1} .(b)\) A combination mercury-steam turbine takes saturated mercury vapor from a boiler at \(469^{\circ} \mathrm{C}\) and exhausts it to heat a steam boiler at \(238^{\circ} \mathrm{C}\). The steam turbine receives steam at this temperature and exhausts it to a condenser at \(37.8^{\circ} \mathrm{C}\). Calculate the maximum efficiency of the combination.

Carnot's theorem states that no heat engine can be more efficient than a Carnot engine (an ideal reversible heat engine) operating between the same temperatures. The efficiency of a heat engine is defined as the ratio of the work done to the heat absorbed - \(\eta = \frac{W}{Q_{\text{absorbed}}}\). For a Carnot engine, this can also be stated in terms of the temperatures of the hot and cold reservoirs - \(\eta_{\text{Carnot}} = 1 - \frac{T_{\text{cold}}}{T_{\text{hot}}}\).

Considering the first part of this problem where a two-stage Carnot engine is described. In the first stage, the engine absorbs heat \(Q_1\) at temperature \(T_1\), does work \(W_1\) and expels heat \(Q_2\) at a lower temperature \(T_2\). By the definition of efficiency, the efficiency of this stage is: \(\frac{W_{1}}{Q_{1}}=1-\frac{T_{2}}{T_{1}}\). For the second stage, it absorbs the heat expelled by the first stage \(Q_2\) at temperature \(T_2\), does work \(W_2\), and expels a quantity of heat \(Q_3\) at a still lower temperature \(T_3\); the efficiency of this second stage can be given by: \(\frac{W_{2}}{Q_{2}} = 1 - \frac{T_{3}}{T_{2}}\). From the first law of thermodynamics, \(Q = W + Q^{\prime}\), where \(Q\) is the absorbed heat, \(W\) is the work done, and \(Q^{\prime}\) is the expelled heat, we can express \(W_2\) in terms of \(Q_1\), \(Q_3\), and \(T_3\) as follows: \(W_{2}=Q_{2}-Q_{3}=Q_{1}-Q_{3}\). Substituting this into the second stage efficiency gives: \(\frac{Q_{1}-Q_{3}}{Q_{1}}=1-\frac{T_{3}}{T_{2}}\), which simplifies to \(\frac{Q_{3}}{Q_{1}}=\frac{T_{3}}{T_{2}}\). Since \(T_2 = T_1\), we can simplify this further to obtain \(Q_{3}=Q_{1} \times \frac{T_{3}}{T_{1}}\). Thus, the total work done \(W\) is \(Q_{1}-Q_{3} = Q_{1} - Q_{1} \times \frac{T_{3}}{T_{1}} = Q_{1} \times (1 - \frac{T_{3}}{T_{1}})\), and therefore the efficiency of the combination, \(\eta = \frac{W}{Q_{\text{absorbed}}} = \frac{Q_{1}(1 - \frac{T_{3}}{T_{1}})}{Q_{1}} = 1 - \frac{T_{3}}{T_{1}} = \frac{T_{1} - T_{3}}{T_{1}}\). Hence we've proved the given equation for the efficiency of a two-stage Carnot heat engine.

Now, calculate the maximum efficiency of a combination mercury-steam turbine using the efficiency formula obtained in Step 2. We convert all the temperatures to the Kelvin scale (as thermal efficiencies must be calculated in absolute temperature) where \(T_{1}=469^{\circ} \mathrm{C} = 469+273.15 = 742.15\mathrm{K}\), \(T_{2}=238^{\circ} \mathrm{C} = 238+273.15 = 511.15\mathrm{K}\), \(T_{3}=37.8^{\circ} \mathrm{C} = 37.8+273.15 = 310.95\mathrm{K}\). Now, by substituting these values into the efficiency formula, we get \(\eta=\frac{T_{1}-T_{3}}{T_{1}}=\frac{742.15-310.95}{742.15}\). Calculating this number will give the maximum efficiency of the combination mercury-steam turbine.