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A Carnot's engine operates between two reservoirs of temperature 900K and 300K. The engine performs 1200 J of work per cycle. The heat energy delivered by the engine to the low temperature reservoir in a cycle is:
Correct answer is '600'. Can you explain this answer?
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A Carnots engine operates between two reservoirs of temperature 900K a...
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Solution:
Given,
- Temperature of the high-temperature reservoir (Th) = 900 K
- Temperature of the low-temperature reservoir (Tl) = 300 K
- Work done per cycle (W) = 1200 J
We need to find the heat energy delivered by the engine to the low-temperature reservoir in a cycle.
We can use the following formula to relate the heat energy supplied to and rejected by the engine to the temperatures of the reservoirs:
Qh / Th = Ql / Tl
where
- Qh is the heat energy supplied to the engine from the high-temperature reservoir
- Ql is the heat energy rejected by the engine to the low-temperature reservoir
From the first law of thermodynamics, we know that
Qh - Ql = W
Substituting the value of Qh / Th from the previous equation, we get:
Ql / Tl = (Qh / Th) - (W / Th)
Substituting the values of Th, Tl, and W, we get:
Ql / 300 = (Qh / 900) - (1200 / 900)
Ql / 300 = (Qh - 4/3)
Multiplying both sides by 300, we get:
Ql = 300(Qh - 4/3)
We know that the Carnot's engine is the most efficient heat engine, and its efficiency is given by:
η = 1 - Tl / Th
Substituting the values of Th and Tl, we get:
η = 1 - (300 / 900) = 2/3
The efficiency of the engine is the ratio of the work done per cycle to the heat energy supplied to the engine from the high-temperature reservoir:
η = W / Qh
Substituting the value of W, we get:
2/3 = 1200 / Qh
Qh = (1200 * 3) / 2 = 1800 J
Substituting the value of Qh in the previous equation, we get:
Ql = 300(1800 - 4/3)
Ql = 539996/3 = 600 J
Therefore, the heat energy delivered by the engine to the low-temperature reservoir in a cycle is 600 J.
Given,
- Temperature of the high-temperature reservoir (Th) = 900 K
- Temperature of the low-temperature reservoir (Tl) = 300 K
- Work done per cycle (W) = 1200 J
We need to find the heat energy delivered by the engine to the low-temperature reservoir in a cycle.
We can use the following formula to relate the heat energy supplied to and rejected by the engine to the temperatures of the reservoirs:
Qh / Th = Ql / Tl
where
- Qh is the heat energy supplied to the engine from the high-temperature reservoir
- Ql is the heat energy rejected by the engine to the low-temperature reservoir
From the first law of thermodynamics, we know that
Qh - Ql = W
Substituting the value of Qh / Th from the previous equation, we get:
Ql / Tl = (Qh / Th) - (W / Th)
Substituting the values of Th, Tl, and W, we get:
Ql / 300 = (Qh / 900) - (1200 / 900)
Ql / 300 = (Qh - 4/3)
Multiplying both sides by 300, we get:
Ql = 300(Qh - 4/3)
We know that the Carnot's engine is the most efficient heat engine, and its efficiency is given by:
η = 1 - Tl / Th
Substituting the values of Th and Tl, we get:
η = 1 - (300 / 900) = 2/3
The efficiency of the engine is the ratio of the work done per cycle to the heat energy supplied to the engine from the high-temperature reservoir:
η = W / Qh
Substituting the value of W, we get:
2/3 = 1200 / Qh
Qh = (1200 * 3) / 2 = 1800 J
Substituting the value of Qh in the previous equation, we get:
Ql = 300(1800 - 4/3)
Ql = 539996/3 = 600 J
Therefore, the heat energy delivered by the engine to the low-temperature reservoir in a cycle is 600 J.
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A Carnots engine operates between two reservoirs of temperature 900K and 300K. The engine performs 1200 J of work per cycle. The heat energy delivered by the engine to the low temperature reservoir in a cycle is:Correct answer is '600'. Can you explain this answer?
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