Just like the efficiency can be used to tell how 'good' an engine is, the Coefficient of Performance can be used to compare refrigerators. If a refrigerator requires 1.21×104 J of work to remove 3.23×104 J of energy from the inside of a refrigerator, what is the coefficient of performance of the refrigerator?
2.How much energy is transferred into the hot reservoir (ie., the kitchen)? Cant figure out how to do this part
3.23x104J / 1.21x104 = 2.66
To determine the coefficient of performance (COP) of a refrigerator, you need to use the formula:
COP = Energy removed (Qc) / Work input (W)
Given that the refrigerator requires 1.21 × 10^4 J of work input and removes 3.23 × 10^4 J of energy, we can calculate the COP as follows:
COP = 3.23 × 10^4 J / 1.21 × 10^4 J
COP = 2.67
Therefore, the coefficient of performance of the refrigerator is 2.67.
For your second question on how much energy is transferred into the hot reservoir (kitchen), we can use the conservation of energy principle. In a refrigerator, the energy removed from the inside (Qc) is equal to the total energy transferred to the hot reservoir (Qh) plus the work input (W).
Qc + W = Qh
In this case, we know the energy removed (Qc) is 3.23 × 10^4 J and the work input (W) is 1.21 × 10^4 J. We can rearrange the equation to solve for Qh:
Qh = Qc + W
Qh = 3.23 × 10^4 J + 1.21 × 10^4 J
Qh = 4.44 × 10^4 J
Therefore, the amount of energy transferred into the hot reservoir (kitchen) is 4.44 × 10^4 J.