A refrigerator has a coefficient of performance equal to 4.2. How much work must be done on the operating gas in the refrigerato
1 answer:
To solve this problem it is necessary to apply the concepts related to the coefficient of performance. The coefficient of performance allows us to relate both the Heat and the Work done on a heat pump or cooling system to know the percentage of losses or input energy.
Mathematically your relationship can be expressed as
Where,
Q = Heat
W = Work
Our values are given as
COP =4.2
Cooling
Replacing at our equation we have,
Therefore the work is 60J.
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Answer:
the rate of flow = 29.28 ×10⁻³ m³/s or 0.029 m³/s
Explanation:
Given:
Diameter of the pipe = 100mm = 0.1m
Contraction ratio = 0.5
thus, diameter at the throat of venturimeter = 0.5×0.1m = 0.05m
The formula for discharge through a venturimeter is given as:
Where,
is the coefficient of discharge = 0.97 (given)
A₁ = Area of the pipe
A₁ =
A₂ = Area at the throat
A₂ =
g = acceleration due to gravity = 9.8m/s²
Now,
The gauge pressure at throat = Absolute pressure - The atmospheric pressure
⇒The gauge pressure at throat = 2 - 10.3 = -8.3 m (Atmosphric pressure = 10.3 m of water)
Thus, the pressure difference at the throat and the pipe = 3- (-8.3) = 11.3m
Substituting the values in the discharge formula we get
or
or
Q = 29.28 ×10⁻³ m³/s
Hence, the rate of flow = 29.28 ×10⁻³ m³/s or 0.029 m³/s