Answer:
W=2 MW
Explanation:
Given that
COP= 2.5
Heat extracted from 85°C
Qa= 5 MW
Lets heat supplied at 150°C = Qr
The power input to heat pump = W
From first law of thermodynamics
Qr= Qa+ W
We know that COP of heat pump given as
W=2 MW
For Carnot heat pump
2.5 T₂ - 895= T₂
T₂=596.66 K
T₂=323.6 °C
Answer:
a) 0.0663 m³/s
b) 3.312 N/(m/s)²
c) 16.665 m/s
d) 0.1105 m³/s
Explanation:
See attached pictures.
Answer:
Equilibrium Temperature is 382.71 K
Total entropy is 0.228 kJ/K
Solution:
As per the question:
Mass of the Aluminium block, M = 28 kg
Initial temperature of aluminium, = 273 + 140 = 413 K
Mass of Iron block, m = 36 kg
Temperature for iron block, = 273 + 60 = 333 K
At 400 k
Specific heat of Aluminium,
At room temperature
Specific heat of iron,
Now,
To calculate the final equilibrium temperature:
Amount of heat loss by Aluminium = Amount of heat gain by Iron
Thus
= 273 + 109.71 = 382.71 K
where
= Equilibrium temperature
Now,
To calculate the changer in entropy:
Now,
For Aluminium:
For Iron:
Thus