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An ideal heat engine operates in Carnot cycle between 227oC and 77oC. It absorbs 2 x 104 cal heat from source at high temperature. The amount of heat rejected to sink at lower temperature is
- a)6 kcal
- b)14 kcal
- c)7 kcal
- d)12 kcal
Correct answer is option 'B'. Can you explain this answer?
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An ideal heat engine operates in Carnot cycle between 227oC and 77oC. ...
Given information:
- The ideal heat engine operates in a Carnot cycle.
- The temperature of the source (high temperature) is 227°C.
- The temperature of the sink (lower temperature) is 77°C.
- The heat absorbed from the source is 2 x 10^4 cal.
Calculating the efficiency of the Carnot cycle:
The efficiency of a Carnot cycle is given by the formula:
Efficiency = 1 - (T_sink / T_source)
Where T_sink is the temperature of the sink and T_source is the temperature of the source.
Substituting the given values:
Efficiency = 1 - (77 + 273) / (227 + 273)
= 1 - 350 / 500
= 1 - 0.7
= 0.3
Calculating the heat rejected to the sink:
The heat rejected to the sink can be calculated using the formula:
Heat rejected = Efficiency * Heat absorbed
Substituting the given values:
Heat rejected = 0.3 * 2 x 10^4 cal
= 6 x 10^3 cal
= 6 kcal
Therefore, the amount of heat rejected to the sink at the lower temperature is 6 kcal, which corresponds to option B.
- The ideal heat engine operates in a Carnot cycle.
- The temperature of the source (high temperature) is 227°C.
- The temperature of the sink (lower temperature) is 77°C.
- The heat absorbed from the source is 2 x 10^4 cal.
Calculating the efficiency of the Carnot cycle:
The efficiency of a Carnot cycle is given by the formula:
Efficiency = 1 - (T_sink / T_source)
Where T_sink is the temperature of the sink and T_source is the temperature of the source.
Substituting the given values:
Efficiency = 1 - (77 + 273) / (227 + 273)
= 1 - 350 / 500
= 1 - 0.7
= 0.3
Calculating the heat rejected to the sink:
The heat rejected to the sink can be calculated using the formula:
Heat rejected = Efficiency * Heat absorbed
Substituting the given values:
Heat rejected = 0.3 * 2 x 10^4 cal
= 6 x 10^3 cal
= 6 kcal
Therefore, the amount of heat rejected to the sink at the lower temperature is 6 kcal, which corresponds to option B.
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An ideal heat engine operates in Carnot cycle between 227oC and 77oC. It absorbs 2 x 104 cal heat from source at high temperature. The amount of heat rejected to sink at lower temperature isa)6 kcalb)14 kcalc)7 kcald)12 kcalCorrect answer is option 'B'. Can you explain this answer?
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An ideal heat engine operates in Carnot cycle between 227oC and 77oC. It absorbs 2 x 104 cal heat from source at high temperature. The amount of heat rejected to sink at lower temperature isa)6 kcalb)14 kcalc)7 kcald)12 kcalCorrect answer is option 'B'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about An ideal heat engine operates in Carnot cycle between 227oC and 77oC. It absorbs 2 x 104 cal heat from source at high temperature. The amount of heat rejected to sink at lower temperature isa)6 kcalb)14 kcalc)7 kcald)12 kcalCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An ideal heat engine operates in Carnot cycle between 227oC and 77oC. It absorbs 2 x 104 cal heat from source at high temperature. The amount of heat rejected to sink at lower temperature isa)6 kcalb)14 kcalc)7 kcald)12 kcalCorrect answer is option 'B'. Can you explain this answer?.
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