Q: A heat pump is a device that absorbs energy from the cold
A heat pump is a device that absorbs energy from the cold outdoor air and transfers it to the warmer indoors. Is this a violation of the second law of thermodynamics? Explain.
See AnswerQ: What is the Clausius expression of the second law of thermodynamics?
What is the Clausius expression of the second law of thermodynamics?
See AnswerQ: Show that the Kelvin–Planck and the Clausius expressions of the
Show that the Kelvin–Planck and the Clausius expressions of the second law are equivalent.
See AnswerQ: The average temperature of the atmosphere in the world is approximated as
The average temperature of the atmosphere in the world is approximated as a function of altitude by the relation Tatm = 288.15 − 6.5z where Tatm is the temperature of the atmosphere in K and z is the...
See AnswerQ: A food freezer is to produce a 5-kW cooling effect
A food freezer is to produce a 5-kW cooling effect, and its COP is 1.3. How many kW of power will this refrigerator require for operation?
See AnswerQ: Describe an imaginary process that satisfies the second law but violates the
Describe an imaginary process that satisfies the second law but violates the first law of thermodynamics.
See AnswerQ: An automotive air conditioner produces a 1-kW cooling effect while
An automotive air conditioner produces a 1-kW cooling effect while consuming 0.75 kW of power. What is the rate at which heat is rejected from this air conditioner?
See AnswerQ: A food refrigerator is to provide a 15,000-kJ
A food refrigerator is to provide a 15,000-kJ/h cooling effect while rejecting 22,000 kJ/h of heat. Calculate the COP of this refrigerator.
See AnswerQ: Determine the COP of a refrigerator that removes heat from the food
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....
See AnswerQ: Determine the COP of a heat pump that supplies energy to a
Determine the COP of a heat pump that supplies energy to a house at a rate of 8000 kJ/h for each kW of electric power it draws. Also, determine the rate of energy absorption from the outdoor air.
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