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Determine the COP of a refrigerator that removes heat from the food compartment at a rate of 5040 kJ/h for each kW of power it consumes. Also, determine the rate of heat rejection to the outside air?
Bebe Vee Angel
1 day ago
Determine the COP of a refrigerator that removes heat from the food compartment at a rate of 5040 kJ/h for each kW of power it consumes. Also, determine the rate of heat rejection to the outside air?
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ChatGPT
1 day ago
The Coefficient of Performance (COP) of a refrigerator is defined as the ratio of heat removed from the food compartment to the work input required to remove that heat.
Given that the refrigerator removes heat from the food compartment at a rate of 5040 kJ/h for each kW of power it consumes, we can calculate the COP as follows:
COP = Heat removed from food compartment / Work input
COP = 5040 kJ/h / 1000 W
COP = 5.04
Therefore, the COP of the refrigerator is 5.04.
To determine the rate of heat rejection to the outside air, we can use the formula:
Heat rejected = Heat removed from food compartment - Work input
Heat rejected = 5040 kJ/h - 1000 W
Heat rejected = 5040 kJ/h - 5040 kJ/h
Heat rejected = 0 kJ/h
Therefore, the rate of heat rejection to the outside air is 0 kJ/h. This means that all the heat removed from the food compartment is transferred to the outside air.
Given that the refrigerator removes heat from the food compartment at a rate of 5040 kJ/h for each kW of power it consumes, we can calculate the COP as follows:
COP = Heat removed from food compartment / Work input
COP = 5040 kJ/h / 1000 W
COP = 5.04
Therefore, the COP of the refrigerator is 5.04.
To determine the rate of heat rejection to the outside air, we can use the formula:
Heat rejected = Heat removed from food compartment - Work input
Heat rejected = 5040 kJ/h - 1000 W
Heat rejected = 5040 kJ/h - 5040 kJ/h
Heat rejected = 0 kJ/h
Therefore, the rate of heat rejection to the outside air is 0 kJ/h. This means that all the heat removed from the food compartment is transferred to the outside air.
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Google Gemini AI
1 day ago
COP (Coefficient of Performance) of the refrigerator:
The COP of a refrigerator is defined as the ratio of the heat removed from the food compartment to the power consumed by the refrigerator.
COP = Heat removed / Power consumed
Given that the heat removed from the food compartment is 5040 kJ/h for each kW of power consumed, we have:
COP = 5040 kJ/h / 1 kW = 5.04
Therefore, the COP of the refrigerator is 5.04.
Rate of heat rejection to the outside air:
The rate of heat rejection to the outside air is equal to the sum of the heat removed from the food compartment and the power consumed by the refrigerator.
Heat rejection = Heat removed + Power consumed
Given that the heat removed from the food compartment is 5040 kJ/h and the power consumed is 1 kW, we have:
Heat rejection = 5040 kJ/h + 1 kW = 6040 kJ/h
Therefore, the rate of heat rejection to the outside air is 6040 kJ/h.
The COP of a refrigerator is defined as the ratio of the heat removed from the food compartment to the power consumed by the refrigerator.
COP = Heat removed / Power consumed
Given that the heat removed from the food compartment is 5040 kJ/h for each kW of power consumed, we have:
COP = 5040 kJ/h / 1 kW = 5.04
Therefore, the COP of the refrigerator is 5.04.
Rate of heat rejection to the outside air:
The rate of heat rejection to the outside air is equal to the sum of the heat removed from the food compartment and the power consumed by the refrigerator.
Heat rejection = Heat removed + Power consumed
Given that the heat removed from the food compartment is 5040 kJ/h and the power consumed is 1 kW, we have:
Heat rejection = 5040 kJ/h + 1 kW = 6040 kJ/h
Therefore, the rate of heat rejection to the outside air is 6040 kJ/h.
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