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A Carnot engine (CE-1) works between two temperature reservoirs A and B, where TA = 900 K and TB = 500 K. A second Carnot engine (CE-2) works between temperature reservoirs B and C, where TC = 300 K. In each cycle of CE-1 and CE-2, all the heat rejected by CE-1 to reservoir B is used by CE-2. For one cycle of operation, if the net Q absorbed by CE-1 from reservoir A is 150 MJ, the net heat rejected to reservoir C by CE-2 (in MJ) is _________.
(Important - Enter only the numerical value in the answer)
Correct answer is '50'. Can you explain this answer?
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A Carnot engine (CE-1) works between two temperature reservoirs A and ...
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A Carnot engine (CE-1) works between two temperature reservoirs A and ...
Given:
CE-1 works between temperature reservoirs A and B, where TA = 900K and TB = 500 K.
CE-2 works between temperature reservoirs B and C, where TC = 300 K.
The net Q absorbed by CE-1 from reservoir A is 150 MJ.
To find:
The net heat rejected to reservoir C by CE-2 (in MJ).
Explanation:
To solve this problem, we can use the concept of the Carnot efficiency and the conservation of energy.
Carnot Efficiency:
The Carnot efficiency of an engine is given by the formula:
η = 1 - (TB/TA)
where η is the efficiency, TB is the lower temperature reservoir, and TA is the higher temperature reservoir.
Using Carnot Efficiency:
1. The efficiency of CE-1 can be calculated as:
η1 = 1 - (TB/TA)
η1 = 1 - (500/900)
η1 = 1 - 0.5556
η1 = 0.4444
2. The efficiency of CE-2 can be calculated as:
η2 = 1 - (TB/TC)
η2 = 1 - (500/300)
η2 = 1 - 1.6667
η2 = -0.6667 (negative value indicates that the engine is a refrigerator and not an engine)
Conservation of Energy:
In a cycle of operation, the net work done by an engine is equal to the net heat absorbed minus the net heat rejected.
Using Conservation of Energy:
For CE-1:
Q1(absorbed) - Q1(rejected) = W1
Q1 - Q1(rejected) = W1
Q1 - Q1(rejected) = η1 * Q1
Q1(rejected) = (1 - η1) * Q1
Q1(rejected) = (1 - 0.4444) * 150 MJ
Q1(rejected) = 0.5556 * 150 MJ
Q1(rejected) = 83.34 MJ
For CE-2:
Q2(rejected) - Q2(absorbed) = W2
Q2(rejected) - Q2 = W2
Q2(rejected) - Q2 = η2 * Q2
Q2(rejected) = (1 - η2) * Q2
Q2(rejected) = (1 - (-0.6667)) * 83.34 MJ
Q2(rejected) = (1 + 0.6667) * 83.34 MJ
Q2(rejected) = 1.6667 * 83.34 MJ
Q2(rejected) = 138.89 MJ
The net heat rejected to reservoir C by CE-2 is 138.89 MJ.
Answer:
The net heat rejected to reservoir C by CE-2 (in MJ) is 138.89.
CE-1 works between temperature reservoirs A and B, where TA = 900K and TB = 500 K.
CE-2 works between temperature reservoirs B and C, where TC = 300 K.
The net Q absorbed by CE-1 from reservoir A is 150 MJ.
To find:
The net heat rejected to reservoir C by CE-2 (in MJ).
Explanation:
To solve this problem, we can use the concept of the Carnot efficiency and the conservation of energy.
Carnot Efficiency:
The Carnot efficiency of an engine is given by the formula:
η = 1 - (TB/TA)
where η is the efficiency, TB is the lower temperature reservoir, and TA is the higher temperature reservoir.
Using Carnot Efficiency:
1. The efficiency of CE-1 can be calculated as:
η1 = 1 - (TB/TA)
η1 = 1 - (500/900)
η1 = 1 - 0.5556
η1 = 0.4444
2. The efficiency of CE-2 can be calculated as:
η2 = 1 - (TB/TC)
η2 = 1 - (500/300)
η2 = 1 - 1.6667
η2 = -0.6667 (negative value indicates that the engine is a refrigerator and not an engine)
Conservation of Energy:
In a cycle of operation, the net work done by an engine is equal to the net heat absorbed minus the net heat rejected.
Using Conservation of Energy:
For CE-1:
Q1(absorbed) - Q1(rejected) = W1
Q1 - Q1(rejected) = W1
Q1 - Q1(rejected) = η1 * Q1
Q1(rejected) = (1 - η1) * Q1
Q1(rejected) = (1 - 0.4444) * 150 MJ
Q1(rejected) = 0.5556 * 150 MJ
Q1(rejected) = 83.34 MJ
For CE-2:
Q2(rejected) - Q2(absorbed) = W2
Q2(rejected) - Q2 = W2
Q2(rejected) - Q2 = η2 * Q2
Q2(rejected) = (1 - η2) * Q2
Q2(rejected) = (1 - (-0.6667)) * 83.34 MJ
Q2(rejected) = (1 + 0.6667) * 83.34 MJ
Q2(rejected) = 1.6667 * 83.34 MJ
Q2(rejected) = 138.89 MJ
The net heat rejected to reservoir C by CE-2 is 138.89 MJ.
Answer:
The net heat rejected to reservoir C by CE-2 (in MJ) is 138.89.
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A Carnot engine (CE-1) works between two temperature reservoirs A and B, where TA = 900K and TB = 500 K. A second Carnot engine (CE-2) works between temperature reservoirs Band C, where TC = 300 K. In each cycle of CE-1 and CE-2, all the heat rejected by CE-1 toreservoir B is used by CE-2. For one cycle of operation, if the net Q absorbed by CE-1 fromreservoir A is 150 MJ, the net heat rejected to reservoir C by CE-2 (in MJ) is _________.(Important - Enter only the numerical value in the answer)Correct answer is '50'. Can you explain this answer?
Question Description
A Carnot engine (CE-1) works between two temperature reservoirs A and B, where TA = 900K and TB = 500 K. A second Carnot engine (CE-2) works between temperature reservoirs Band C, where TC = 300 K. In each cycle of CE-1 and CE-2, all the heat rejected by CE-1 toreservoir B is used by CE-2. For one cycle of operation, if the net Q absorbed by CE-1 fromreservoir A is 150 MJ, the net heat rejected to reservoir C by CE-2 (in MJ) is _________.(Important - Enter only the numerical value in the answer)Correct answer is '50'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about A Carnot engine (CE-1) works between two temperature reservoirs A and B, where TA = 900K and TB = 500 K. A second Carnot engine (CE-2) works between temperature reservoirs Band C, where TC = 300 K. In each cycle of CE-1 and CE-2, all the heat rejected by CE-1 toreservoir B is used by CE-2. For one cycle of operation, if the net Q absorbed by CE-1 fromreservoir A is 150 MJ, the net heat rejected to reservoir C by CE-2 (in MJ) is _________.(Important - Enter only the numerical value in the answer)Correct answer is '50'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A Carnot engine (CE-1) works between two temperature reservoirs A and B, where TA = 900K and TB = 500 K. A second Carnot engine (CE-2) works between temperature reservoirs Band C, where TC = 300 K. In each cycle of CE-1 and CE-2, all the heat rejected by CE-1 toreservoir B is used by CE-2. For one cycle of operation, if the net Q absorbed by CE-1 fromreservoir A is 150 MJ, the net heat rejected to reservoir C by CE-2 (in MJ) is _________.(Important - Enter only the numerical value in the answer)Correct answer is '50'. Can you explain this answer?.
A Carnot engine (CE-1) works between two temperature reservoirs A and B, where TA = 900K and TB = 500 K. A second Carnot engine (CE-2) works between temperature reservoirs Band C, where TC = 300 K. In each cycle of CE-1 and CE-2, all the heat rejected by CE-1 toreservoir B is used by CE-2. For one cycle of operation, if the net Q absorbed by CE-1 fromreservoir A is 150 MJ, the net heat rejected to reservoir C by CE-2 (in MJ) is _________.(Important - Enter only the numerical value in the answer)Correct answer is '50'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about A Carnot engine (CE-1) works between two temperature reservoirs A and B, where TA = 900K and TB = 500 K. A second Carnot engine (CE-2) works between temperature reservoirs Band C, where TC = 300 K. In each cycle of CE-1 and CE-2, all the heat rejected by CE-1 toreservoir B is used by CE-2. For one cycle of operation, if the net Q absorbed by CE-1 fromreservoir A is 150 MJ, the net heat rejected to reservoir C by CE-2 (in MJ) is _________.(Important - Enter only the numerical value in the answer)Correct answer is '50'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A Carnot engine (CE-1) works between two temperature reservoirs A and B, where TA = 900K and TB = 500 K. A second Carnot engine (CE-2) works between temperature reservoirs Band C, where TC = 300 K. In each cycle of CE-1 and CE-2, all the heat rejected by CE-1 toreservoir B is used by CE-2. For one cycle of operation, if the net Q absorbed by CE-1 fromreservoir A is 150 MJ, the net heat rejected to reservoir C by CE-2 (in MJ) is _________.(Important - Enter only the numerical value in the answer)Correct answer is '50'. Can you explain this answer?.
Solutions for A Carnot engine (CE-1) works between two temperature reservoirs A and B, where TA = 900K and TB = 500 K. A second Carnot engine (CE-2) works between temperature reservoirs Band C, where TC = 300 K. In each cycle of CE-1 and CE-2, all the heat rejected by CE-1 toreservoir B is used by CE-2. For one cycle of operation, if the net Q absorbed by CE-1 fromreservoir A is 150 MJ, the net heat rejected to reservoir C by CE-2 (in MJ) is _________.(Important - Enter only the numerical value in the answer)Correct answer is '50'. Can you explain this answer? in English & in Hindi are available as part of our courses for GATE.
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A Carnot engine (CE-1) works between two temperature reservoirs A and B, where TA = 900K and TB = 500 K. A second Carnot engine (CE-2) works between temperature reservoirs Band C, where TC = 300 K. In each cycle of CE-1 and CE-2, all the heat rejected by CE-1 toreservoir B is used by CE-2. For one cycle of operation, if the net Q absorbed by CE-1 fromreservoir A is 150 MJ, the net heat rejected to reservoir C by CE-2 (in MJ) is _________.(Important - Enter only the numerical value in the answer)Correct answer is '50'. Can you explain this answer?, a detailed solution for A Carnot engine (CE-1) works between two temperature reservoirs A and B, where TA = 900K and TB = 500 K. A second Carnot engine (CE-2) works between temperature reservoirs Band C, where TC = 300 K. In each cycle of CE-1 and CE-2, all the heat rejected by CE-1 toreservoir B is used by CE-2. For one cycle of operation, if the net Q absorbed by CE-1 fromreservoir A is 150 MJ, the net heat rejected to reservoir C by CE-2 (in MJ) is _________.(Important - Enter only the numerical value in the answer)Correct answer is '50'. Can you explain this answer? has been provided alongside types of A Carnot engine (CE-1) works between two temperature reservoirs A and B, where TA = 900K and TB = 500 K. A second Carnot engine (CE-2) works between temperature reservoirs Band C, where TC = 300 K. In each cycle of CE-1 and CE-2, all the heat rejected by CE-1 toreservoir B is used by CE-2. For one cycle of operation, if the net Q absorbed by CE-1 fromreservoir A is 150 MJ, the net heat rejected to reservoir C by CE-2 (in MJ) is _________.(Important - Enter only the numerical value in the answer)Correct answer is '50'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice A Carnot engine (CE-1) works between two temperature reservoirs A and B, where TA = 900K and TB = 500 K. A second Carnot engine (CE-2) works between temperature reservoirs Band C, where TC = 300 K. In each cycle of CE-1 and CE-2, all the heat rejected by CE-1 toreservoir B is used by CE-2. For one cycle of operation, if the net Q absorbed by CE-1 fromreservoir A is 150 MJ, the net heat rejected to reservoir C by CE-2 (in MJ) is _________.(Important - Enter only the numerical value in the answer)Correct answer is '50'. Can you explain this answer? tests, examples and also practice GATE tests.
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