A carnot refrigerator uses an ideal carnot cycle to cool its interior. The outside air has a temp of 304K and the refrigerator is kept at a temp of 279K.
1) what is the coefficient of performance of this refrigerator?
2) if the refrigerator does 1500J of work on the working substance, how much energy is transferred out of the refrigerator's interior?
3) in the scenario described above, how much energy is transferred into outside air ?
To answer these questions, we need to understand the concept of Carnot refrigerators and the Carnot cycle.
1) The coefficient of performance (COP) of a refrigerator is defined as the ratio of the amount of heat removed from the cooling space (Qc) to the work done on the refrigerant (W). Mathematically, the COP is given by: COP = Qc / W.
In a Carnot cycle, the COP can be calculated using the formula:
COP = Tc / (Th - Tc) where Tc is the absolute temperature of the cold reservoir (in this case, the inside of the refrigerator) and Th is the absolute temperature of the hot reservoir (the outside air).
Using the given temperatures:
Tc = 279K
Th = 304K
COP = 279 / (304 - 279) = 279 / 25 = 11.16
Therefore, the coefficient of performance of this refrigerator is 11.16.
2) The amount of energy transferred out of the refrigerator's interior can be calculated using the first law of thermodynamics, which states that energy cannot be created or destroyed, only transferred or transformed. In this case, the work done on the working substance is 1500J.
Since work is defined as the energy transferred by a force acting through a distance, we can say that the work done is equal to the energy transferred out of the system.
So, the energy transferred out of the refrigerator's interior is 1500J.
3) In the scenario described above, the energy transferred into the outside air can be determined using the first law of thermodynamics again. Since energy cannot be created or destroyed, the energy transferred out of the refrigerator's interior must be equal to the energy transferred into the outside air.
Therefore, the energy transferred into the outside air is also 1500J.
To solve these questions, we'll use the equations related to the Carnot cycle and the coefficient of performance (COP) of a refrigerator.
1) The coefficient of performance (COP) of a refrigerator is given by the equation:
COP = Qc / W,
where Qc is the amount of heat removed from the cold reservoir, and W is the work done on the working substance.
In this case, we are given the temperatures:
Tc (temperature of the cold reservoir) = 279K,
Th (temperature of the hot reservoir or outside air) = 304K.
The COP of the refrigerator can be calculated as follows:
COP = Th / (Th - Tc)
COP = 304K / (304K - 279K)
COP = 304K / 25K
COP = 12.16
So, the coefficient of performance of this refrigerator is 12.16.
2) To calculate the amount of energy transferred out of the refrigerator's interior, we need to use the equation:
Qc = W - Qin,
where Qin is the amount of heat transferred into the refrigerator's interior.
In this case, we are given the work done on the working substance:
W = 1500J
Substituting the given values into the equation, we get:
1500J = Qc - Qin
Since we want to calculate the amount of energy transferred out of the refrigerator's interior (Qc), rearrange the equation:
Qc = 1500J + Qin
So, the amount of energy transferred out of the refrigerator's interior is 1500J + Qin.
3) In a Carnot refrigerator, the heat transferred to the cold reservoir (Qc) is equal to the amount of heat rejected to the outside air. Therefore, to determine the energy transferred into the outside air, we can use the equation:
Q_rejected = Qc
So, the amount of energy transferred into the outside air is Q_rejected = Qc.