(II) (a) Given that the coefficient of performance of a...

(II) $(a)$ Given that the coefficient of performance of a refrigerator is defined (Eq. 4a) as $$\mathrm{COP}=\frac{Q_{\mathrm{L}}}{W}$$ show that for an ideal (Carnot) refrigerator, (b) Write the COP in terms of the efficiency e of the reversible heat engine obtained by running the refrigerator backward. (c) What is the coefficient of performance for an ideal refrigerator that maintains a freezer compartment at $-18^{\circ} \mathrm{C}$ when the condenser's temperature is $24^{\circ} \mathrm{C} ?$

(II) $(a)$ Given that the coefficient of performance of a refrigerator is defined (Eq. 4a) as
$$\mathrm{COP}=\frac{Q_{\mathrm{L}}}{W}$$ show that for an ideal (Carnot) refrigerator,
(b) Write the COP in terms of the efficiency e of the reversible heat engine obtained by running the refrigerator backward. (c) What is the coefficient of performance for an ideal refrigerator that maintains a freezer compartment at $-18^{\circ} \mathrm{C}$ when the condenser's temperature is $24^{\circ} \mathrm{C} ?$

Key Concepts

Coefficient of Performance (COP)

The COP is a measure of the effectiveness of a refrigerator, defined as the ratio of the heat removed from the low-temperature reservoir (Q_L) to the work (W) input needed to remove that heat. It is a key metric in thermodynamics that helps assess the performance of cooling devices, especially when comparing ideal (reversible) and real (irreversible) systems.

Carnot Cycle

The Carnot cycle represents an idealized reversible cycle that sets the upper limit on the efficiency of any heat engine or refrigerator working between two temperature reservoirs. It serves as the benchmark for real-world systems because it involves reversible processes, meaning that there is no net entropy production, and any deviation from this cycle results in lower performance.

Reversible Heat Engines and Efficiency

In thermodynamics, a reversible heat engine is an idealized system that converts heat into work or vice versa without any losses due to irreversibility. The efficiency (e) of a reversible heat engine is defined by the ratio of work output to heat input. When such an engine is run in reverse, it acts as a refrigerator, and its coefficient of performance is directly related to the original efficiency, illustrating the fundamental connection between the two operating modes.

Relationship Between Efficiency and COP

For systems based on reversible processes, there is a clear mathematical relationship between the efficiency of a heat engine and the coefficient of performance of the refrigerator derived from operating the engine in reverse. This relationship stems from the conservation of energy and the second law of thermodynamics, and it allows the transformation of one performance metric into the other using the temperatures of the heat reservoirs.

Temperature Scales in Thermodynamics

Thermodynamic analyses require the use of an absolute temperature scale, typically Kelvin, to ensure that all relationships, such as those involving efficiency and COP, are correctly applied. Conversion from Celsius to Kelvin is critical when using formulas derived from the Carnot cycle, as the temperature differences and ratios must be computed in absolute terms to maintain accuracy.

Transcript

00:01 Prove that the coefficient of performance of an ideal engine is going to be equal to the lower temperature, the lower temperature divided by the higher temperature minus the lower temperature, referring to the operating temperatures of this refrigerator.

00:20 So we can say that the coefficient of performance of an ideal engine is going to be equal to the heat exhausted q sub l, divided by the work done.

00:33 And we can say that this is going to be equal to the heat exhausted divided by q sub h.

00:40 This would be the heat input minus heat exhaust.

00:45 We're going to divide all of these terms by q subh.

00:52 So we can say that this is going to be q sub h divided by q sub h minus q sub l divided by q subh once again and then here q sub l divided by q sub h.

01:10 At this point we can say that this is going to be equal to for the denominator it'll be the higher temperature divided by the higher temperature minus the lower temperature divided by the higher temperature and then for the numerator the lower temperature divided by the higher temperature.

01:39 And at this point, we can eliminate t sub h.

01:42 And this is going to equal the lower temperature divided by the difference between the higher temperature and the lower temperature.

01:52 So this would be our final solution for part a.

01:56 This is how we would prove the coefficient of the equation for the coefficient of performance of an ideal refrigerator...

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