Post

Created by @nathanedwards
 at October 31st 2023, 8:54:26 pm.

Question:

A heat engine operates between two reservoirs at temperatures of 400 K and 800 K. During each cycle, the engine absorbs 300 J of heat from the high-temperature reservoir, expels 200 J of heat to the low-temperature reservoir, and performs work. Assume the engine is operating in an idealized Carnot cycle.

a) Calculate the efficiency of this heat engine.

b) Determine the amount of work done by the engine during each cycle.

c) If the engine operates for 1000 cycles, calculate the total heat absorbed from the high-temperature reservoir during this process.

d) Is this heat engine operating in accordance with the first law of thermodynamics? Explain your answer.

Answer:

a) The efficiency of a heat engine is given by the formula:

η=1TlowThigh \eta = 1 - \frac{{T_{\text{{low}}}}}{{T_{\text{{high}}}}}

where η\eta is the efficiency and TlowT_{\text{{low}}} and ThighT_{\text{{high}}} are the temperatures of the low and high-temperature reservoirs, respectively.

Given: Thigh=800KT_{\text{{high}}} = 800 \, \text{{K}} and Tlow=400KT_{\text{{low}}} = 400 \, \text{{K}}

Substituting the values into the formula, we have:

η=1400K800K0.5 \eta = 1 - \frac{{400 \, \text{{K}}}}{{800 \, \text{{K}}}} \approx 0.5

Therefore, the efficiency of the heat engine is approximately 0.5 or 50%.

b) The work done by a heat engine is given by the formula:

W=QinQout W = Q_{\text{{in}}} - Q_{\text{{out}}}

where WW is the work done, QinQ_{\text{{in}}} is the heat absorbed from the high-temperature reservoir, and QoutQ_{\text{{out}}} is the heat expelled to the low-temperature reservoir.

Given: Qin=300JQ_{\text{{in}}} = 300 \, \text{{J}} and Qout=200JQ_{\text{{out}}} = 200 \, \text{{J}}

Substituting the values into the formula, we have:

W=300J200J=100J W = 300 \, \text{{J}} - 200 \, \text{{J}} = 100 \, \text{{J}}

Therefore, the engine does 100 J of work during each cycle.

c) The total heat absorbed from the high-temperature reservoir during 1000 cycles can be calculated by multiplying the heat absorbed during each cycle by the number of cycles:

Total heat absorbed=Qin×number of cycles \text{{Total heat absorbed}} = Q_{\text{{in}}} \times \text{{number of cycles}}

Given: Qin=300JQ_{\text{{in}}} = 300 \, \text{{J}}, number of cycles=1000\text{{number of cycles}} = 1000

Substituting the values into the formula, we have:

Total heat absorbed=300J×1000=300000J \text{{Total heat absorbed}} = 300 \, \text{{J}} \times 1000 = 300000 \, \text{{J}}

Therefore, the total heat absorbed from the high-temperature reservoir during this process is 300,000 J.

d) Yes, this heat engine is operating in accordance with the first law of thermodynamics. The first law states that energy cannot be created or destroyed, only transferred or transformed. In this case, the heat engine absorbs 300 J of heat from the high-temperature reservoir, expels 200 J of heat to the low-temperature reservoir, and performs 100 J of work during each cycle. The total energy input (300 J) plus the total energy output (200 J) equals the total work done (100 J). Therefore, energy is conserved, and the heat engine is operating in accordance with the first law of thermodynamics.