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Consider a refrigerator with a COP of 1.2. It removes heat from the refrigerated space at a rate of 60 kJ/min. What will be the electric power consumed by the refrigerator and the rate of heat transfer to the kitchen air, respectively?
- a)1.04 kW, 60 kJ/min
- b)0.83 kW, 110 kJ/min
- c)1.25 kW, 50 kJ/min
- d)kW, 93.5 kJ/min
Correct answer is option 'B'. Can you explain this answer?
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Most Upvoted Answer
Consider a refrigerator with a COP of 1.2. It removes heat from the re...
Using the definition of the coefficient of performance, the power z input to the refrigerator is determined to be
The heat transfer rate to the kitchen air is determined from the energy balance as
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Consider a refrigerator with a COP of 1.2. It removes heat from the re...
Given Data:
Refrigeration rate = 60 kJ/min
COP = 1.2
Calculations:
Electric Power Consumed:
Electric power consumed by the refrigerator can be calculated using the formula:
\[COP = \frac{Refrigeration \ Rate}{Electric \ Power \ Consumed}\]
\[1.2 = \frac{60}{Electric \ Power \ Consumed}\]
\[Electric \ Power \ Consumed = \frac{60}{1.2} = 50 \ kW\]
Rate of Heat Transfer to Kitchen Air:
The rate of heat transfer to the kitchen air can be calculated using the formula:
\[COP = \frac{Refrigeration \ Rate}{Electric \ Power \ Consumed}\]
Rate of heat transfer to kitchen air = Refrigeration rate + Electric power consumed
\[Rate \ of \ heat \ transfer \ to \ kitchen \ air = 60 + 50 = 110 \ kJ/min\]
Therefore, the electric power consumed by the refrigerator is 0.83 kW and the rate of heat transfer to the kitchen air is 110 kJ/min. Hence, option B is the correct answer.
Refrigeration rate = 60 kJ/min
COP = 1.2
Calculations:
Electric Power Consumed:
Electric power consumed by the refrigerator can be calculated using the formula:
\[COP = \frac{Refrigeration \ Rate}{Electric \ Power \ Consumed}\]
\[1.2 = \frac{60}{Electric \ Power \ Consumed}\]
\[Electric \ Power \ Consumed = \frac{60}{1.2} = 50 \ kW\]
Rate of Heat Transfer to Kitchen Air:
The rate of heat transfer to the kitchen air can be calculated using the formula:
\[COP = \frac{Refrigeration \ Rate}{Electric \ Power \ Consumed}\]
Rate of heat transfer to kitchen air = Refrigeration rate + Electric power consumed
\[Rate \ of \ heat \ transfer \ to \ kitchen \ air = 60 + 50 = 110 \ kJ/min\]
Therefore, the electric power consumed by the refrigerator is 0.83 kW and the rate of heat transfer to the kitchen air is 110 kJ/min. Hence, option B is the correct answer.
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Consider a refrigerator with a COP of 1.2. It removes heat from the refrigerated space at a rate of 60 kJ/min. What will be the electric power consumed by the refrigerator and the rate of heat transfer to the kitchen air, respectively?a)1.04 kW, 60 kJ/minb)0.83 kW, 110 kJ/minc)1.25 kW, 50 kJ/mind)kW, 93.5 kJ/minCorrect answer is option 'B'. Can you explain this answer?
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Consider a refrigerator with a COP of 1.2. It removes heat from the refrigerated space at a rate of 60 kJ/min. What will be the electric power consumed by the refrigerator and the rate of heat transfer to the kitchen air, respectively?a)1.04 kW, 60 kJ/minb)0.83 kW, 110 kJ/minc)1.25 kW, 50 kJ/mind)kW, 93.5 kJ/minCorrect answer is option 'B'. 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 Consider a refrigerator with a COP of 1.2. It removes heat from the refrigerated space at a rate of 60 kJ/min. What will be the electric power consumed by the refrigerator and the rate of heat transfer to the kitchen air, respectively?a)1.04 kW, 60 kJ/minb)0.83 kW, 110 kJ/minc)1.25 kW, 50 kJ/mind)kW, 93.5 kJ/minCorrect answer is option 'B'. 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 Consider a refrigerator with a COP of 1.2. It removes heat from the refrigerated space at a rate of 60 kJ/min. What will be the electric power consumed by the refrigerator and the rate of heat transfer to the kitchen air, respectively?a)1.04 kW, 60 kJ/minb)0.83 kW, 110 kJ/minc)1.25 kW, 50 kJ/mind)kW, 93.5 kJ/minCorrect answer is option 'B'. Can you explain this answer?.
Consider a refrigerator with a COP of 1.2. It removes heat from the refrigerated space at a rate of 60 kJ/min. What will be the electric power consumed by the refrigerator and the rate of heat transfer to the kitchen air, respectively?a)1.04 kW, 60 kJ/minb)0.83 kW, 110 kJ/minc)1.25 kW, 50 kJ/mind)kW, 93.5 kJ/minCorrect answer is option 'B'. 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 Consider a refrigerator with a COP of 1.2. It removes heat from the refrigerated space at a rate of 60 kJ/min. What will be the electric power consumed by the refrigerator and the rate of heat transfer to the kitchen air, respectively?a)1.04 kW, 60 kJ/minb)0.83 kW, 110 kJ/minc)1.25 kW, 50 kJ/mind)kW, 93.5 kJ/minCorrect answer is option 'B'. 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 Consider a refrigerator with a COP of 1.2. It removes heat from the refrigerated space at a rate of 60 kJ/min. What will be the electric power consumed by the refrigerator and the rate of heat transfer to the kitchen air, respectively?a)1.04 kW, 60 kJ/minb)0.83 kW, 110 kJ/minc)1.25 kW, 50 kJ/mind)kW, 93.5 kJ/minCorrect answer is option 'B'. Can you explain this answer?.
Solutions for Consider a refrigerator with a COP of 1.2. It removes heat from the refrigerated space at a rate of 60 kJ/min. What will be the electric power consumed by the refrigerator and the rate of heat transfer to the kitchen air, respectively?a)1.04 kW, 60 kJ/minb)0.83 kW, 110 kJ/minc)1.25 kW, 50 kJ/mind)kW, 93.5 kJ/minCorrect answer is option 'B'. Can you explain this answer? in English & in Hindi are available as part of our courses for Mechanical Engineering.
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Consider a refrigerator with a COP of 1.2. It removes heat from the refrigerated space at a rate of 60 kJ/min. What will be the electric power consumed by the refrigerator and the rate of heat transfer to the kitchen air, respectively?a)1.04 kW, 60 kJ/minb)0.83 kW, 110 kJ/minc)1.25 kW, 50 kJ/mind)kW, 93.5 kJ/minCorrect answer is option 'B'. Can you explain this answer?, a detailed solution for Consider a refrigerator with a COP of 1.2. It removes heat from the refrigerated space at a rate of 60 kJ/min. What will be the electric power consumed by the refrigerator and the rate of heat transfer to the kitchen air, respectively?a)1.04 kW, 60 kJ/minb)0.83 kW, 110 kJ/minc)1.25 kW, 50 kJ/mind)kW, 93.5 kJ/minCorrect answer is option 'B'. Can you explain this answer? has been provided alongside types of Consider a refrigerator with a COP of 1.2. It removes heat from the refrigerated space at a rate of 60 kJ/min. What will be the electric power consumed by the refrigerator and the rate of heat transfer to the kitchen air, respectively?a)1.04 kW, 60 kJ/minb)0.83 kW, 110 kJ/minc)1.25 kW, 50 kJ/mind)kW, 93.5 kJ/minCorrect answer is option 'B'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice Consider a refrigerator with a COP of 1.2. It removes heat from the refrigerated space at a rate of 60 kJ/min. What will be the electric power consumed by the refrigerator and the rate of heat transfer to the kitchen air, respectively?a)1.04 kW, 60 kJ/minb)0.83 kW, 110 kJ/minc)1.25 kW, 50 kJ/mind)kW, 93.5 kJ/minCorrect answer is option 'B'. Can you explain this answer? tests, examples and also practice Mechanical Engineering tests.
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