Question

A heat engine receives heat from a heat source at \(1200^{\circ} \mathrm{C}\) and has a thermal efficiency of 40 percent. The heat engine does maximum work equal to \(500 \mathrm{kJ}\). Determine the heat supplied to the heat engine by the heat source, the heat rejected to the heat sink, and the temperature of the heat \(\sin \mathrm{k}\)

Short answer

Question: Determine the heat supplied by the heat source, the heat rejected to the heat sink, and the temperature of the heat sink for the given heat engine. Answer: The heat supplied by the heat source is \(1250\,\mathrm{kJ}\), the heat rejected to the heat sink is \(750\,\mathrm{kJ}\), and the temperature of the heat sink is \(610.74^{\circ}\mathrm{C}\).

Step-by-step solution

Convert temperature to Kelvin scale

First, convert the temperature of the heat source from Celsius to Kelvin: \(T_1 = 1200^{\circ}\mathrm{C} + 273.15 \mathrm{K} = 1473.15\mathrm{K}\)

Determine the heat supplied by the heat source

We'll use the given thermal efficiency and the maximum work done by the engine to find the heat supplied: \(\eta = \frac{W}{Q_1}\) We are given \(\eta = 0.40\) and \(W = 500\,\mathrm{kJ}\). We can rearrange the formula for \(Q_1\): \(Q_1 = \frac{W}{\eta} = \frac{500\,\mathrm{kJ}}{0.40} = 1250\,\mathrm{kJ}\)

Determine the heat rejected to the heat sink

Using the first law of thermodynamics for a heat engine, we have: \(W = Q_1 - Q_2\) We can solve for \(Q_2\): \(Q_2 = Q_1 - W = 1250\,\mathrm{kJ} - 500\,\mathrm{kJ} = 750\,\mathrm{kJ}\)

Determine the temperature of the heat sink

Now let's use the Carnot efficiency formula to find the temperature of the heat sink: Carnot Efficiency = \(1 - \frac{T_2}{T_1}\) Since the engine is operating at its maximum efficiency, we can assume that the given thermal efficiency is equal to the Carnot efficiency. Therefore: \(0.40 = 1-\frac{T_2}{1473.15}\) Solve for \(T_2\): \(T_2 = 1473.15\mathrm{K} \times (1-0.40) = 1473.15\mathrm{K} \times 0.60 = 883.89\,\mathrm{K}\)

Convert the temperature of the heat sink to Celsius scale

Finally, convert the temperature of the heat sink from Kelvin to Celsius: \(T_2 = 883.89\,\mathrm{K} - 273.15\,\mathrm{K} = 610.74^{\circ}\mathrm{C}\) To summarize: Heat supplied by the heat source: \(Q_1 = 1250\,\mathrm{kJ}\) Heat rejected to the heat sink: \(Q_2 = 750\,\mathrm{kJ}\) Temperature of the heat sink: \(T_2 = 610.74 ^{\circ}\mathrm{C}\)