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A refrigerator works b/w -3 degree cel. and 27 degree cel. The work done by the compressor for each joule of heat absorbed from the freezer is (i)9Joule ii)10 J iii) 1/9 J (iv)1/10 J?
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A refrigerator works b/w -3 degree cel. and 27 degree cel. The work do...
Refrigerator Efficiency Calculation
In order to determine the work done by the compressor for each joule of heat absorbed from the freezer, we can use the formula for the coefficient of performance (COP) of a refrigerator. The COP is defined as the ratio of the heat extracted from the cold reservoir to the work input to the system.
Given Data:
- Temperature of the freezer (Tc) = -3 degree Celsius = 270 K
- Maximum temperature of the refrigerator (Th) = 27 degree Celsius = 300 K
Calculating COP:
COP = Th / (Th - Tc)
COP = 300 / (300 - 270)
COP = 300 / 30
COP = 10
Understanding Work Done:
The work done by the compressor for each joule of heat absorbed from the freezer can be calculated using the formula:
Work done = Heat absorbed / COP
Since we are interested in the work done for each joule of heat absorbed, we can consider the heat absorbed to be 1 Joule.
Calculating Work Done:
Work done = 1 / 10
Work done = 0.1 Joule
Therefore, the correct answer is (iv) 1/10 J. The compressor does 0.1 Joule of work for each Joule of heat absorbed from the freezer.
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A refrigerator works b/w -3 degree cel. and 27 degree cel. The work do...
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A refrigerator works b/w -3 degree cel. and 27 degree cel. The work done by the compressor for each joule of heat absorbed from the freezer is (i)9Joule ii)10 J iii) 1/9 J (iv)1/10 J?
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