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If a refrigerator is used for heating purposes in winter so that the atmosphere becomes the cold body
and the room to be heated becomes the hot body, how much heat would be available for heating for
each kW input to the driving motor? The COP of the refrigerator is 5, and the electromechanical
efficiency of the motor is 90%. How does this compare with resistance heating?
and the room to be heated becomes the hot body, how much heat would be available for heating for
each kW input to the driving motor? The COP of the refrigerator is 5, and the electromechanical
efficiency of the motor is 90%. How does this compare with resistance heating?
Most Upvoted Answer
If a refrigerator is used for heating purposes in winter so that the a...
Refrigerator used for heating purposes in winter
COP Calculation
- COP (Coefficient of Performance) = Heat Output / Work Input
- COP = 5 (given)
- Therefore, Heat Output = 5 * Work Input
Electromechanical Efficiency
- Electromechanical Efficiency = Output Power / Input Power
- Electromechanical Efficiency = 90% (given)
- Therefore, Output Power = 0.9 * Input Power
Heat Available for Heating
- Heat Available for Heating = Heat Output - Work Input
- Heat Available for Heating = 5 * Work Input - Input Power
- Heat Available for Heating = Input Power * (5 - 1/0.9)
- Heat Available for Heating = Input Power * 4.12
Comparison with Resistance Heating
- Resistance Heating Efficiency = 100% (as all input power is converted to heat)
- Heat Available for Heating (Resistance Heating) = Input Power
- Therefore, Heat Available for Heating (Refrigerator) / Heat Available for Heating (Resistance Heating) = 4.12
Explanation
The COP of a refrigerator is a measure of its efficiency in transferring heat from a cold body to a hot body. In this case, the cold body is the atmosphere and the hot body is the room to be heated. The COP of the refrigerator is given as 5, which means that for every unit of work input, the refrigerator produces 5 units of heat output. However, not all of this heat output is available for heating, as some of it is lost due to inefficiencies in the electromechanical system. The electromechanical efficiency of the motor is given as 90%, which means that only 90% of the input power is converted to output power. Therefore, the heat available for heating is equal to the heat output minus the input power, which is equivalent to the input power multiplied by 4.12.
When compared to resistance heating, where all input power is converted to heat, the heat available for heating from the refrigerator is only 41.2% of the input power. Therefore, resistance heating is much more efficient for heating purposes compared to using a refrigerator.
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If a refrigerator is used for heating purposes in winter so that the atmosphere becomes the cold body and the room to be heated becomes the hot body, how much heat would be available for heating for each kW input to the driving motor? The COP of the refrigerator is 5, and the electromechanical efficiency of the motor is 90%. How does this compare with resistance heating?
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If a refrigerator is used for heating purposes in winter so that the atmosphere becomes the cold body and the room to be heated becomes the hot body, how much heat would be available for heating for each kW input to the driving motor? The COP of the refrigerator is 5, and the electromechanical efficiency of the motor is 90%. How does this compare with resistance heating? for Software Development 2025 is part of Software Development preparation. The Question and answers have been prepared according to the Software Development exam syllabus. Information about If a refrigerator is used for heating purposes in winter so that the atmosphere becomes the cold body and the room to be heated becomes the hot body, how much heat would be available for heating for each kW input to the driving motor? The COP of the refrigerator is 5, and the electromechanical efficiency of the motor is 90%. How does this compare with resistance heating? covers all topics & solutions for Software Development 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If a refrigerator is used for heating purposes in winter so that the atmosphere becomes the cold body and the room to be heated becomes the hot body, how much heat would be available for heating for each kW input to the driving motor? The COP of the refrigerator is 5, and the electromechanical efficiency of the motor is 90%. How does this compare with resistance heating?.
If a refrigerator is used for heating purposes in winter so that the atmosphere becomes the cold body and the room to be heated becomes the hot body, how much heat would be available for heating for each kW input to the driving motor? The COP of the refrigerator is 5, and the electromechanical efficiency of the motor is 90%. How does this compare with resistance heating? for Software Development 2025 is part of Software Development preparation. The Question and answers have been prepared according to the Software Development exam syllabus. Information about If a refrigerator is used for heating purposes in winter so that the atmosphere becomes the cold body and the room to be heated becomes the hot body, how much heat would be available for heating for each kW input to the driving motor? The COP of the refrigerator is 5, and the electromechanical efficiency of the motor is 90%. How does this compare with resistance heating? covers all topics & solutions for Software Development 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If a refrigerator is used for heating purposes in winter so that the atmosphere becomes the cold body and the room to be heated becomes the hot body, how much heat would be available for heating for each kW input to the driving motor? The COP of the refrigerator is 5, and the electromechanical efficiency of the motor is 90%. How does this compare with resistance heating?.
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If a refrigerator is used for heating purposes in winter so that the atmosphere becomes the cold body and the room to be heated becomes the hot body, how much heat would be available for heating for each kW input to the driving motor? The COP of the refrigerator is 5, and the electromechanical efficiency of the motor is 90%. How does this compare with resistance heating?, a detailed solution for If a refrigerator is used for heating purposes in winter so that the atmosphere becomes the cold body and the room to be heated becomes the hot body, how much heat would be available for heating for each kW input to the driving motor? The COP of the refrigerator is 5, and the electromechanical efficiency of the motor is 90%. How does this compare with resistance heating? has been provided alongside types of If a refrigerator is used for heating purposes in winter so that the atmosphere becomes the cold body and the room to be heated becomes the hot body, how much heat would be available for heating for each kW input to the driving motor? The COP of the refrigerator is 5, and the electromechanical efficiency of the motor is 90%. How does this compare with resistance heating? theory, EduRev gives you an
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