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Chapter 15: Problem 42

Show that the coefficient of performance for a reversible heat pump is $1 /\left(1-T_{\mathrm{C}} / T_{\mathrm{H}}\right)$.

Short Answer

Expert verified
Question: Show that the coefficient of performance (COP) for a reversible heat pump is given by the formula: \[COP = \frac{1}{1-\frac{T_C}{T_H}}\]

Step by step solution

01

Define the coefficient of performance for a heat pump

The coefficient of performance (COP) for a heat pump is the ratio of useful heat energy transferred to the cold reservoir (Q_C) to the work input (W) required to run the heat pump. Mathematically, it is given by: \[COP = \frac{Q_C}{W}\]
02

Define the efficiency of a reversible heat engine

The efficiency of a heat engine is the ratio of the work output (W) to the heat energy input (Q_H). A reversible heat engine has the maximum efficiency possible and the efficiency is given by: \[\eta_{max} = 1 - \frac{T_C}{T_H}\]
03

Express work (W) in terms of heat energies (Q_H and Q_C)

For a reversible heat engine, we have the principle of energy conservation, meaning that the total energy balance is given by: \[Q_H = Q_C + W\] Now, we can solve for W: \[W = Q_H - Q_C\]
04

Substitute the relation of W in COP equation

Replace W in the COP equation with the expression we found in Step 3: \[COP = \frac{Q_C}{Q_H - Q_C}\]
05

Express Q_C in terms of Q_H and T_C and T_H

From the efficiency equation, we can get the relationship between heat energies and temperatures as follows: \[\eta_{max} = 1 - \frac{T_C}{T_H} = 1 - \frac{Q_C}{Q_H}\] Now, we can solve for Q_C: \[Q_C = Q_H \frac{T_C}{T_H}\]
06

Substitute the relation of Q_C in the COP equation

Replace Q_C in the COP equation with the expression we found in Step 5: \[COP = \frac{Q_H \frac{T_C}{T_H}}{Q_H - Q_H \frac{T_C}{T_H}}\]
07

Simplify the COP equation

Now, we can simplify the equation to get the desired COP formula: \[\begin{aligned} COP &= \frac{Q_H \frac{T_C}{T_H}}{Q_H \left(1 - \frac{T_C}{T_H}\right)} \\ &= \frac{\frac{T_C}{T_H}}{1 - \frac{T_C}{T_H}} \\ &= \frac{1}{\frac{1}{(\frac{T_C}{T_H})} - 1} \\ &= \frac{1}{1-\frac{T_C}{T_H}} \end{aligned}\] So, the coefficient of performance for a reversible heat pump is: \[COP = \frac{1}{1-\frac{T_C}{T_H}}\]

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Most popular questions from this chapter

Within an insulated system, \(418.6 \mathrm{kJ}\) of heat is conducted through a copper rod from a hot reservoir at \(+200.0^{\circ} \mathrm{C}\) to a cold reservoir at \(+100.0^{\circ} \mathrm{C} .\) (The reservoirs are so big that this heat exchange does not change their temperatures appreciably.) What is the net change in entropy of the system, in \(\mathrm{kJ} / \mathrm{K} ?\)
(a) Calculate the efficiency of a reversible engine that operates between the temperatures \(600.0^{\circ} \mathrm{C}\) and \(300.0^{\circ} \mathrm{C} .\) (b) If the engine absorbs \(420.0 \mathrm{kJ}\) of heat from the hot reservoir, how much does it exhaust to the cold reservoir?
In a heat engine, 3.00 mol of a monatomic ideal gas, initially at 4.00 atm of pressure, undergoes an isothermal expansion, increasing its volume by a factor of 9.50 at a constant temperature of \(650.0 \mathrm{K} .\) The gas is then compressed at a constant pressure to its original volume. Finally, the pressure is increased at constant volume back to the original pressure. (a) Draw a \(P V\) diagram of this three-step heat engine. (b) For each step of this process, calculate the work done on the gas, the change in internal energy, and the heat transferred into the gas. (c) What is the efficiency of this engine?
The efficiency of a muscle during weight lifting is equal to the work done in lifting the weight divided by the total energy output of the muscle (work done plus internal energy dissipated in the muscle). Determine the efficiency of a muscle that lifts a \(161-\mathrm{N}\) weight through a vertical displacement of \(0.577 \mathrm{m}\) and dissipates \(139 \mathrm{J}\) in the process.
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