Answer:
a. 2.08, b. 1110 kJ/min
Explanation:
The power consumption and the cooling rate of an air conditioner are given. The COP or Coefficient of Performance and the rate of heat rejection are to be determined. <u>Assume that the air conditioner operates steadily.</u>
a. The coefficient of performance of the air conditioner (refrigerator) is determined from its definition, which is
COP(r) = Q(L)/W(net in), where Q(L) is the rate of heat removed and W(net in) is the work done to remove said heat
COP(r) = (750 kJ/min/6 kW) x (1 kW/60kJ/min) = 2.08
The COP of this air conditioner is 2.08.
b. The rate of heat discharged to the outside air is determined from the energy balance.
Q(H) = Q(L) + W(net in)
Q(H) = 750 kJ/min + 6 x 60 kJ/min = 1110 kJ/min
The rate of heat transfer to the outside air is 1110 kJ for every minute.
I would go with C but i am not 100 percent on that
Answer: the answer will be d because it is the right one to be
Explanation:
Answer:
T=151 K, U=-1.848*10^6J
Explanation:
The given process occurs when the pressure is constant. Given gas follows the Ideal Gas Law:
pV=nRT
For the given scenario, we operate with the amount of the gas- n- calculated in moles. To find n, we use molar mass: M=102 g/mol.
Using the given mass m, molar mass M, we can get the following equation:
pV=mRT/M
To calculate change in the internal energy, we need to know initial and final temperatures. We can calculate both temperatures as:
T=pVM/(Rm); so initial T=302.61K and final T=151.289K
Now we can calculate change of U:
U=3/2 mRT/M using T- difference in temperatures
U=-1.848*10^6 J
Note, that the energy was taken away from the system.
