Question

A heat engine receives heat from a source at \(1000^{\circ} \mathrm{C}\) and rejects the waste heat to a sink at \(50^{\circ} \mathrm{C}\). If heat is supplied to this engine at a rate of \(100 \mathrm{kJ} / \mathrm{s}\), the maximum power this heat engine can produce is (a) \(25.4 \mathrm{kW}\) (b) \(55.4 \mathrm{kW}\) \((c) 74.6 \mathrm{kW}\) \((d) 95.0 \mathrm{kW}\) \((e) 100 \mathrm{kW}\)

Short answer

Answer: The maximum power produced by the heat engine is 74.6 kW.

Step-by-step solution

Convert temperatures to Kelvin

Before calculating the efficiency, we need to convert the temperatures from Celsius to Kelvin by adding 273.15 to each: $$ T_1 = 1000 + 273.15 = 1273.15 \thinspace K \newline T_2 = 50 + 273.15 = 323.15 \thinspace K $$

Calculate efficiency for a Carnot engine

The Carnot efficiency is given by the formula: $$ \eta = 1 - \frac{T_2}{T_1} $$ Substitute the temperatures calculated above: $$ \eta = 1 - \frac{323.15 \thinspace K}{1273.15 \thinspace K} = 1 - 0.2538 = 0.7462 $$

Calculate the maximum power

The maximum power can be calculated by using the Carnot efficiency and the rate at which heat is supplied: $$ P_\text{max} = \eta \times Q_\text{input} \newline P_\text{max} = 0.7462 \times 100 \thinspace \frac{\text{kJ}}{\text{s}} $$ Convert the heat input rate to watts: $$ Q_\text{input} = 100 \thinspace \frac{\text{kJ}}{\text{s}} = 100000 \thinspace \frac{\text{J}}{\text{s}} = 100000 \thinspace W $$ Now, find the maximum power: $$ P_\text{max} = 0.7462 \times 100000 \thinspace W = 74600 \thinspace W = 74.6 \thinspace kW $$ The maximum power this heat engine can produce is 74.6 kW, which corresponds to option (c).