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A heat engine having an efficiency of 70% is used to drive a refrigerator having a coefficient of performance of 5. The energy absorbed from low temperature reservoir by the refrigerator for each kJ of energy absorbed from high temperature source by the engine is
- a)0.14kJ
- b)0.71kJ
- c)3.5 kJ
- d)7.1 kJ
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
A heat engine having an efficiency of 70% is used to drive a refrigera...
Given: ηengine = 0.7, (C.O.P)R = 5
...(i)
Now,
...(ii)
Again,
Energy absorbed from low temperature reservoir by the refrigerator for each kJ of energy absorbed from high temperature source by the engine = 3.5 kJ.
...(i)Now,
...(ii)Again,
Energy absorbed from low temperature reservoir by the refrigerator for each kJ of energy absorbed from high temperature source by the engine = 3.5 kJ.
Most Upvoted Answer
A heat engine having an efficiency of 70% is used to drive a refrigera...
Efficiency of the Heat Engine
Given that the efficiency of the heat engine is 70%, we can represent it mathematically as:
Efficiency = (Useful work output) / (Energy input)
In this case, the useful work output is the energy absorbed from the high-temperature source, and the energy input is that absorbed from the low-temperature reservoir. Let's represent the energy absorbed from the high-temperature source as Qh, and the energy absorbed from the low-temperature reservoir as Ql.
Therefore, the equation for efficiency becomes:
0.70 = Qh / Ql
Coefficient of Performance of the Refrigerator
The coefficient of performance (COP) of a refrigerator is defined as the ratio of the heat extracted from the low-temperature reservoir (Ql) to the work input (W) required to operate the refrigerator. Mathematically, it can be represented as:
COP = Ql / W
In this case, the COP is given as 5, which means that for every unit of work input, the refrigerator extracts 5 units of heat from the low-temperature reservoir.
Finding the Energy Absorbed from the Low-Temperature Reservoir
To find the energy absorbed from the low-temperature reservoir for each kilojoule (kJ) of energy absorbed from the high-temperature source, we need to relate the efficiency of the heat engine and the COP of the refrigerator.
From the equations for efficiency and COP, we can rearrange them as follows:
Qh = 0.70 * Ql
Ql = 5 * W
Substituting the value of Ql from the second equation into the first equation, we get:
Qh = 0.70 * (5 * W)
Simplifying further, we have:
Qh = 3.5 * W
Since Qh represents the energy absorbed from the high-temperature source for each kilojoule of energy absorbed from the low-temperature reservoir, the answer is 3.5 kJ (option C).
Given that the efficiency of the heat engine is 70%, we can represent it mathematically as:
Efficiency = (Useful work output) / (Energy input)
In this case, the useful work output is the energy absorbed from the high-temperature source, and the energy input is that absorbed from the low-temperature reservoir. Let's represent the energy absorbed from the high-temperature source as Qh, and the energy absorbed from the low-temperature reservoir as Ql.
Therefore, the equation for efficiency becomes:
0.70 = Qh / Ql
Coefficient of Performance of the Refrigerator
The coefficient of performance (COP) of a refrigerator is defined as the ratio of the heat extracted from the low-temperature reservoir (Ql) to the work input (W) required to operate the refrigerator. Mathematically, it can be represented as:
COP = Ql / W
In this case, the COP is given as 5, which means that for every unit of work input, the refrigerator extracts 5 units of heat from the low-temperature reservoir.
Finding the Energy Absorbed from the Low-Temperature Reservoir
To find the energy absorbed from the low-temperature reservoir for each kilojoule (kJ) of energy absorbed from the high-temperature source, we need to relate the efficiency of the heat engine and the COP of the refrigerator.
From the equations for efficiency and COP, we can rearrange them as follows:
Qh = 0.70 * Ql
Ql = 5 * W
Substituting the value of Ql from the second equation into the first equation, we get:
Qh = 0.70 * (5 * W)
Simplifying further, we have:
Qh = 3.5 * W
Since Qh represents the energy absorbed from the high-temperature source for each kilojoule of energy absorbed from the low-temperature reservoir, the answer is 3.5 kJ (option C).
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A heat engine having an efficiency of 70% is used to drive a refrigerator having a coefficient of performance of 5. The energy absorbed from low temperature reservoir by the refrigerator for each kJ of energy absorbed from high temperature source by the engine isa)0.14kJb)0.71kJc)3.5 kJd)7.1 kJCorrect answer is option 'C'. Can you explain this answer?
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A heat engine having an efficiency of 70% is used to drive a refrigerator having a coefficient of performance of 5. The energy absorbed from low temperature reservoir by the refrigerator for each kJ of energy absorbed from high temperature source by the engine isa)0.14kJb)0.71kJc)3.5 kJd)7.1 kJCorrect answer is option 'C'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about A heat engine having an efficiency of 70% is used to drive a refrigerator having a coefficient of performance of 5. The energy absorbed from low temperature reservoir by the refrigerator for each kJ of energy absorbed from high temperature source by the engine isa)0.14kJb)0.71kJc)3.5 kJd)7.1 kJCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A heat engine having an efficiency of 70% is used to drive a refrigerator having a coefficient of performance of 5. The energy absorbed from low temperature reservoir by the refrigerator for each kJ of energy absorbed from high temperature source by the engine isa)0.14kJb)0.71kJc)3.5 kJd)7.1 kJCorrect answer is option 'C'. Can you explain this answer?.
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