A Carnot engine is operated between two heat reservoirs at temperatures 520 K and 300 K. (a) If the engine receives 6.45 kJ of heat energy from the reservoir at 520 K in each cycle, how many joules per cycle does it reject to the reservoir at 300 K? (b) How much mechanical work is performed by the engine during each cycle? (c) What is the thermal efficiency of the engine?
To calculate the answers to these questions, we can use the formulas and principles of the Carnot cycle. The Carnot cycle is an idealized thermodynamic cycle that consists of four processes:
1. Isothermal Expansion (A to B): In this process, the working substance (gas) expands and absorbs heat from the high-temperature reservoir (520 K) while maintaining a constant temperature.
2. Adiabatic Expansion (B to C): The working substance continues to expand without exchanging heat with the surroundings. During this process, the temperature decreases.
3. Isothermal Compression (C to D): The working substance is compressed, and heat is rejected to the low-temperature reservoir (300 K) while maintaining a constant temperature.
4. Adiabatic Compression (D to A): The substance is compressed further without exchanging heat.
For part (a) of the question, we need to calculate the amount of heat rejected to the low-temperature reservoir. Since the Carnot cycle is reversible, the amount of heat rejected during the isothermal compression process is equal to the amount of heat absorbed during the isothermal expansion process.
To find this, we can use the formula for the efficiency of a Carnot engine:
Efficiency = 1 - (Temperature of the Low-Temperature Reservoir / Temperature of the High-Temperature Reservoir)
Efficiency = 1 - (300 K / 520 K)
Efficiency = 1 - 0.577
Efficiency = 0.423
This means that the Carnot engine converts only 42.3% of the heat energy it receives into useful work, while the remaining 57.7% is rejected to the low-temperature reservoir.
To calculate the amount of heat rejected, we can multiply the efficiency by the heat energy received from the high-temperature reservoir:
Heat Rejected = Efficiency * Heat Energy Received
Heat Rejected = 0.423 * 6.45 kJ
Heat Rejected ≈ 2.73 kJ
Therefore, the engine rejects approximately 2.73 kJ of heat to the low-temperature reservoir in each cycle.
For part (b) of the question, we need to calculate the mechanical work performed by the engine. The work done by the engine in each cycle is equal to the difference between the heat energy absorbed and the heat energy rejected.
Mechanical Work = Heat Energy Received - Heat Energy Rejected
Mechanical Work = 6.45 kJ - 2.73 kJ
Mechanical Work ≈ 3.72 kJ
Therefore, the engine performs approximately 3.72 kJ of mechanical work during each cycle.
For part (c) of the question, we need to calculate the thermal efficiency of the engine. The thermal efficiency is defined as the ratio of the mechanical work done by the engine to the heat energy received from the high-temperature reservoir.
Thermal Efficiency = (Mechanical Work / Heat Energy Received) * 100%
Thermal Efficiency = (3.72 kJ / 6.45 kJ) * 100%
Thermal Efficiency ≈ 57.6%
Therefore, the thermal efficiency of the engine is approximately 57.6%.