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A carnot engine operating between two temperatures 727°C and 27°C is supplied heat energy at the rate of 500 joule/cycle. 60% of the work output is used to derive a refrigerator, which rejects heat to the surrounding at 27°C. If the refrigerator removes 1050 Joule of heat per cycle from the low temperature reservoir, determine the temperature (in Kelvin) of reservoir.
Correct answer is '250'. Can you explain this answer?
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Verified Answer
A carnot engine operating between two temperatures 727°C and 27&de...
Efficiency 
Work output
In refrigeration Q1 + W = Q2
Work input = 60% of work output of heat engine
Work input = 0.6 * 350 = 210 J
Heat removed Q1 = 1050 J
Heat rejected Q2 = Q1 + W = 1050 + 210 = 1260J
Now,
Work output
In refrigeration Q1 + W = Q2
Work input = 60% of work output of heat engine
Work input = 0.6 * 350 = 210 J
Heat removed Q1 = 1050 J
Heat rejected Q2 = Q1 + W = 1050 + 210 = 1260J
Now,
Most Upvoted Answer
A carnot engine operating between two temperatures 727°C and 27&de...
Efficiency 
Work output
In refrigeration Q1 + W = Q2
Work input = 60% of work output of heat engine
Work input = 0.6 * 350 = 210 J
Heat removed Q1 = 1050 J
Heat rejected Q2 = Q1 + W = 1050 + 210 = 1260J
Now,
Work output
In refrigeration Q1 + W = Q2
Work input = 60% of work output of heat engine
Work input = 0.6 * 350 = 210 J
Heat removed Q1 = 1050 J
Heat rejected Q2 = Q1 + W = 1050 + 210 = 1260J
Now,
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Community Answer
A carnot engine operating between two temperatures 727°C and 27&de...
°C and 27°C has an efficiency of 50%. If the engine absorbs 10,000 J of heat energy from the hot reservoir, calculate the heat energy rejected to the cold reservoir.
The efficiency of a Carnot engine is given by the formula:
Efficiency = 1 - (Tc/Th),
where Tc is the temperature of the cold reservoir and Th is the temperature of the hot reservoir.
Given that the efficiency is 50% (0.50) and the temperature of the hot reservoir (Th) is 727°C, we can rearrange the formula to solve for Tc:
0.50 = 1 - (Tc/727).
Simplifying the equation:
0.50 = 1 - Tc/727.
Rearranging:
Tc/727 = 1 - 0.50,
Tc/727 = 0.50.
Multiplying both sides by 727:
Tc = 0.50 * 727,
Tc = 363.5°C.
Now, we can calculate the heat energy rejected to the cold reservoir using the equation:
Heat energy rejected = Efficiency * Heat energy absorbed.
Given that the heat energy absorbed is 10,000 J and the efficiency is 0.50, we can calculate:
Heat energy rejected = 0.50 * 10,000 J,
Heat energy rejected = 5,000 J.
Therefore, the heat energy rejected to the cold reservoir is 5,000 J.
The efficiency of a Carnot engine is given by the formula:
Efficiency = 1 - (Tc/Th),
where Tc is the temperature of the cold reservoir and Th is the temperature of the hot reservoir.
Given that the efficiency is 50% (0.50) and the temperature of the hot reservoir (Th) is 727°C, we can rearrange the formula to solve for Tc:
0.50 = 1 - (Tc/727).
Simplifying the equation:
0.50 = 1 - Tc/727.
Rearranging:
Tc/727 = 1 - 0.50,
Tc/727 = 0.50.
Multiplying both sides by 727:
Tc = 0.50 * 727,
Tc = 363.5°C.
Now, we can calculate the heat energy rejected to the cold reservoir using the equation:
Heat energy rejected = Efficiency * Heat energy absorbed.
Given that the heat energy absorbed is 10,000 J and the efficiency is 0.50, we can calculate:
Heat energy rejected = 0.50 * 10,000 J,
Heat energy rejected = 5,000 J.
Therefore, the heat energy rejected to the cold reservoir is 5,000 J.
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A carnot engine operating between two temperatures 727°C and 27°C is supplied heat energy at the rate of 500 joule/cycle. 60% of the work output is used to derive a refrigerator, which rejects heat to the surrounding at 27°C. If the refrigerator removes 1050 Joule of heat per cycle from the low temperature reservoir, determine the temperature (in Kelvin) of reservoir.Correct answer is '250'. Can you explain this answer?
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A carnot engine operating between two temperatures 727°C and 27°C is supplied heat energy at the rate of 500 joule/cycle. 60% of the work output is used to derive a refrigerator, which rejects heat to the surrounding at 27°C. If the refrigerator removes 1050 Joule of heat per cycle from the low temperature reservoir, determine the temperature (in Kelvin) of reservoir.Correct answer is '250'. Can you explain this answer? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about A carnot engine operating between two temperatures 727°C and 27°C is supplied heat energy at the rate of 500 joule/cycle. 60% of the work output is used to derive a refrigerator, which rejects heat to the surrounding at 27°C. If the refrigerator removes 1050 Joule of heat per cycle from the low temperature reservoir, determine the temperature (in Kelvin) of reservoir.Correct answer is '250'. Can you explain this answer? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A carnot engine operating between two temperatures 727°C and 27°C is supplied heat energy at the rate of 500 joule/cycle. 60% of the work output is used to derive a refrigerator, which rejects heat to the surrounding at 27°C. If the refrigerator removes 1050 Joule of heat per cycle from the low temperature reservoir, determine the temperature (in Kelvin) of reservoir.Correct answer is '250'. Can you explain this answer?.
A carnot engine operating between two temperatures 727°C and 27°C is supplied heat energy at the rate of 500 joule/cycle. 60% of the work output is used to derive a refrigerator, which rejects heat to the surrounding at 27°C. If the refrigerator removes 1050 Joule of heat per cycle from the low temperature reservoir, determine the temperature (in Kelvin) of reservoir.Correct answer is '250'. Can you explain this answer? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about A carnot engine operating between two temperatures 727°C and 27°C is supplied heat energy at the rate of 500 joule/cycle. 60% of the work output is used to derive a refrigerator, which rejects heat to the surrounding at 27°C. If the refrigerator removes 1050 Joule of heat per cycle from the low temperature reservoir, determine the temperature (in Kelvin) of reservoir.Correct answer is '250'. Can you explain this answer? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A carnot engine operating between two temperatures 727°C and 27°C is supplied heat energy at the rate of 500 joule/cycle. 60% of the work output is used to derive a refrigerator, which rejects heat to the surrounding at 27°C. If the refrigerator removes 1050 Joule of heat per cycle from the low temperature reservoir, determine the temperature (in Kelvin) of reservoir.Correct answer is '250'. Can you explain this answer?.
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