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A heat pump working on the Carnot cycle maintains the inside temperature of a house at 22ºC by supplying 450 kJ/s. If the outside temperature is 0ºC, the heat taken, in kJ/s, from the outside air is approximately.
Correct answer is '416.44'. Can you explain this answer?
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A heat pump working on the Carnot cycle maintains the inside temperatu...
The efficiency of Carnot heat engine may be defined as
Q1 = 450kJ/s, Q2 = ?
T1 = 22 + 273 = 295K
T2 = 273K
Hence,
= 416.44 kJ/s
The correct answer is: 416.44
Q1 = 450kJ/s, Q2 = ?
T1 = 22 + 273 = 295K
T2 = 273K
Hence,
= 416.44 kJ/s
The correct answer is: 416.44
Most Upvoted Answer
A heat pump working on the Carnot cycle maintains the inside temperatu...
°C when the outside temperature is -5°C. The heat pump consumes 3 kW of power. Determine the rate of heat transfer to the house, the coefficient of performance of the heat pump, and the rate of heat transfer from the outside to the heat pump.
To determine the rate of heat transfer to the house, we first need to calculate the heat pump's coefficient of performance (COP) using the Carnot cycle formula:
COP = Th / (Th - Tc)
Where Th is the temperature at which heat is transferred to the house and Tc is the temperature at which heat is absorbed from the outside.
Given:
Th = 22°C
Tc = -5°C
Plugging these values into the formula, we have:
COP = 22 / (22 - (-5))
COP = 22 / 27
COP ≈ 0.815
The coefficient of performance (COP) of the heat pump is approximately 0.815.
Now, to determine the rate of heat transfer to the house, we use the formula:
Qh = COP * W
Where Qh is the rate of heat transfer to the house and W is the power consumed by the heat pump.
Given:
COP ≈ 0.815
W = 3 kW
Plugging the values into the formula, we have:
Qh = 0.815 * 3
Qh ≈ 2.445 kW
The rate of heat transfer to the house is approximately 2.445 kW.
Finally, to determine the rate of heat transfer from the outside to the heat pump, we can use the formula:
Qc = Qh - W
Where Qc is the rate of heat transfer from the outside to the heat pump.
Given:
Qh ≈ 2.445 kW
W = 3 kW
Plugging the values into the formula, we have:
Qc = 2.445 - 3
Qc ≈ -0.555 kW
The rate of heat transfer from the outside to the heat pump is approximately -0.555 kW. This negative value indicates that heat is being extracted from the outside to heat the house.
To determine the rate of heat transfer to the house, we first need to calculate the heat pump's coefficient of performance (COP) using the Carnot cycle formula:
COP = Th / (Th - Tc)
Where Th is the temperature at which heat is transferred to the house and Tc is the temperature at which heat is absorbed from the outside.
Given:
Th = 22°C
Tc = -5°C
Plugging these values into the formula, we have:
COP = 22 / (22 - (-5))
COP = 22 / 27
COP ≈ 0.815
The coefficient of performance (COP) of the heat pump is approximately 0.815.
Now, to determine the rate of heat transfer to the house, we use the formula:
Qh = COP * W
Where Qh is the rate of heat transfer to the house and W is the power consumed by the heat pump.
Given:
COP ≈ 0.815
W = 3 kW
Plugging the values into the formula, we have:
Qh = 0.815 * 3
Qh ≈ 2.445 kW
The rate of heat transfer to the house is approximately 2.445 kW.
Finally, to determine the rate of heat transfer from the outside to the heat pump, we can use the formula:
Qc = Qh - W
Where Qc is the rate of heat transfer from the outside to the heat pump.
Given:
Qh ≈ 2.445 kW
W = 3 kW
Plugging the values into the formula, we have:
Qc = 2.445 - 3
Qc ≈ -0.555 kW
The rate of heat transfer from the outside to the heat pump is approximately -0.555 kW. This negative value indicates that heat is being extracted from the outside to heat the house.
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A heat pump working on the Carnot cycle maintains the inside temperature of a house at 22ºC by supplying 450 kJ/s. If the outside temperature is 0ºC, the heat taken, in kJ/s, from the outside air is approximately.Correct answer is '416.44'. Can you explain this answer?
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A heat pump working on the Carnot cycle maintains the inside temperature of a house at 22ºC by supplying 450 kJ/s. If the outside temperature is 0ºC, the heat taken, in kJ/s, from the outside air is approximately.Correct answer is '416.44'. Can you explain this answer? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about A heat pump working on the Carnot cycle maintains the inside temperature of a house at 22ºC by supplying 450 kJ/s. If the outside temperature is 0ºC, the heat taken, in kJ/s, from the outside air is approximately.Correct answer is '416.44'. Can you explain this answer? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A heat pump working on the Carnot cycle maintains the inside temperature of a house at 22ºC by supplying 450 kJ/s. If the outside temperature is 0ºC, the heat taken, in kJ/s, from the outside air is approximately.Correct answer is '416.44'. Can you explain this answer?.
A heat pump working on the Carnot cycle maintains the inside temperature of a house at 22ºC by supplying 450 kJ/s. If the outside temperature is 0ºC, the heat taken, in kJ/s, from the outside air is approximately.Correct answer is '416.44'. Can you explain this answer? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about A heat pump working on the Carnot cycle maintains the inside temperature of a house at 22ºC by supplying 450 kJ/s. If the outside temperature is 0ºC, the heat taken, in kJ/s, from the outside air is approximately.Correct answer is '416.44'. Can you explain this answer? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A heat pump working on the Carnot cycle maintains the inside temperature of a house at 22ºC by supplying 450 kJ/s. If the outside temperature is 0ºC, the heat taken, in kJ/s, from the outside air is approximately.Correct answer is '416.44'. Can you explain this answer?.
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