What is the best coefficient of performance for a refrigerator that cools an environment at −27.0°C and has heat transfer to another environment at 49.5°C?
To determine the best coefficient of performance (COP) for a refrigerator, we need to use the formula:
COP = Desired Cooling/Work Input
In this case, the desired cooling is the amount of heat transferred from the environment at -27.0°C to the environment at 49.5°C. To find this value, we can use the formula:
Qcool = m * c * ΔT
Where:
Qcool : Desired cooling (heat transferred)
m : Mass of the substance being cooled
c : Specific heat capacity of the substance
ΔT : Change in temperature (final temperature - initial temperature)
First, we need to convert -27.0°C and 49.5°C to Kelvin (K) since temperature values should be in Kelvin for calculations:
T1 (initial temperature) = -27.0°C + 273.15 = 246.15 K
T2 (final temperature) = 49.5°C + 273.15 = 322.65 K
Now we can calculate the change in temperature (ΔT):
ΔT = T2 - T1 = 322.65 K - 246.15 K = 76.5 K
Next, we need to consider the specific heat capacity (c) of the substance being cooled. This value depends on the material, and it is usually provided in J/(kg·K).
Assuming the specific heat capacity is c = 4186 J/(kg·K), and let's assume the mass of the substance being cooled is 1 kg, we can calculate the desired cooling (Qcool):
Qcool = m * c * ΔT = 1 kg * 4186 J/(kg·K) * 76.5 K = 320,471 J
Now we have the desired cooling (Qcool). To find the work input (W), we need to convert the units to joules (J):
1 Watt-hour (Wh) = 3600 J
Let's assume the refrigerator runs for 1 hour. The work input (W) in joules is equal to the power input (P) multiplied by the time (t):
P = W / t
Assuming the refrigerator consumes 300 Watts (W) of power:
W = P * t = 300 W * 1 hour * 3600 J = 1,080,000 J
Now we can find the best coefficient of performance (COP) using the formula:
COP = Qcool / W
COP = 320,471 J / 1,080,000 J
COP ≈ 0.297
Therefore, the best coefficient of performance (COP) for the given refrigerator is approximately 0.297.