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?
For a Carnot engine,
Qin/Qout = Tin/Tout = 520/300 = 1.733
Wout = Qin - Qout = Qin[1 - Qout/Qin]
= Qin[1 - (Tout/Tin)] = ___
Efficiency = Wout/Qin = __
.935
An ideal heat engine operates between two temperature 200.5k and 769.1k what is the effi of the engine
To answer these questions, we need to use the laws of thermodynamics and the formula for the efficiency of a Carnot engine.
The efficiency of a Carnot engine is given by the formula:
Efficiency = (Th - Tc) / Th
where Th is the temperature of the hot reservoir and Tc is the temperature of the cold reservoir.
(a) To calculate how much heat energy the engine rejects to the reservoir at 300 K, we can use the formula:
Qc = Efficiency * Qh
Given that Qh = 6.45 kJ and Th = 520 K, we can solve for Qc:
Qc = (300 / 520) * 6.45 kJ
Qc ≈ 3.716 kJ
So, the engine rejects approximately 3.716 kJ of heat energy to the reservoir at 300 K in each cycle.
(b) The mechanical work performed by the engine during each cycle can be calculated using the formula:
W = Qh - Qc
Given that Qh = 6.45 kJ and Qc = 3.716 kJ, we can find W:
W = 6.45 kJ - 3.716 kJ
W ≈ 2.734 kJ
Therefore, the engine performs approximately 2.734 kJ of mechanical work during each cycle.
(c) The thermal efficiency of the engine can be calculated using the formula mentioned earlier:
Efficiency = (Th - Tc) / Th
Given that Th = 520 K and Tc = 300 K, we can find the efficiency:
Efficiency = (520 - 300) / 520
Efficiency ≈ 0.423
So, the thermal efficiency of the engine is approximately 42.3%.