Respuesta :
Answer:
a) Seems true at first glance, but on further inspection, the statement is false.
b) True
c) False
d) True
e) True
f) False
g) False
Explanation:
Taking the statements one by one
a.) The Clausius statement denies the possibility of heat transfer from a cooler to a hotter body.
The Clausius Statement denies the possibility of heat transfer from a cooler to a hotter body without extra work. It does not outrightly state that there is no possibility of heat transfer from a cooler to a hotter body.
For example, an Air conditioner or refrigerator rejects heat from a cold reservoir to a hot reservoir.
So, this statement is false.
b) The COP of a reversible refrigeration cycles is equal or greater than the COP of an irreversible refrigeration cycle if both cycles operate between the same thermal reservoirs.
The Coefficient of Performance of a reversible cycle is the maximum efficiency possible. It is the efficiency of a Carnot Engine.
Hence, it is greater than or equal to the Coefficient of Performance of an irreversible cycle.
COP(reversible) ≥ COP(irreversible)
This statement is true.
c. Mass, energy, and temperature are the examples of intensive properties.
Intensive properties are properties of thermodynamic systems that do not depend on the extent of the system. They are the same for a particular size of substance and stay the same if the size of the substance is doubled or halved. Examples include temperature, specific capacity, specific volume, every specific property basically, etc.
Extensive properties depend on the extent of the system. They double or half when the size of the extent doubles or halves respectively.
Mass and Energy are Extensive properties.
Temperature is the only intensive property among these options.
This statement is false.
d. For reversible refrigeration and heat pump cycles operating between the same hot and cold reservoirs, the relation between their coefficients of performance is
COPHP = COPR + 1.
The coefficients of performance for reversible refrigeration and heat pump cycles operating between the same hot and cold reservoirs are indeed related through
COP(HP) = COP(R) + 1
COP of a heat pump = COP(HP) = (Qh/W)
COP of a refrigerator = COP(R) = (Qc/W)
But, Qh = Qc + W
Divide through by W
(Qh/W) = (Qc/W) + (W/W)
COP(HP) = COP(R) + 1 (Proved!)
This Statement is true.
e. For a reversible heat pump that operates between cold and hot thermal reservoirs at 350°C and 550°C, respectively, the COP is equal to 4.11.
The COP is given as (1/efficiency).
Efficiency = 1 - (Tc/Th)
Tc = temperature of cold thermal reservoir in Kelvin = 350°C = 623.15 K
Th = temperature of hot thermal reservoir in Kelvin = 550°C = 823.15 K
Efficiency = 1 - (623.15/823.15)
= 1 - 0.757 = 0.243
COP = (1/Efficiency) = (1/0.243) = 4.11
This statement is true.
f. In the absence of any friction and other irreversibilities, a heat engine can achieve an efficiency of 100%.
In the absence of friction and other irreversibilities and for a heat engine to have 100% efficiency, the temperature of its cold reservoir has to be 0 K or the tempersture of its hot reservoir has to be infinity.
Efficiency = 1 - (Tc/Th)
For efficiency to be 1,
(Tc/Th) = 0; that is, Tc = 0 or Th = infinity
These two aren't physically possible and for 100% efficiency to happen, the heat engine will have to violate the Kelvin-Planck's statement of the second law of thermodynamics.
According to Kelvin-Planck's statement of the second law of thermodynamics, net amount of work cannot be produced by exchanging heat with single reservoir i.e. there will be another reservoir to reject heat.
Hence, a heat engine cannot have an efficiency of 100%.
This statement is false.
g. A refrigerator, with a COP of 1.2, rejects 60 kJ/min from a refrigerated space when the electric power consumed by the refrigerator is 50 kJ/min. This refrigerator violates the first law of thermodynamics.
The COP of a refrigerator is given as
COP = (Qcold)/W
Qcold = Heat rejected from the cold reservoir = Heat rejected from refrigerated space = 60 KJ/min
W = work done on the system = electrical power consumed = 50 KJ/min
COP = (60/50) = 1.2
This system does not violate the first law of thermodynamics.
This statement is false.
Hope this Helps!!!
