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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

The United States generates about \(5.0 \times 10^{16} \mathrm{J}\) of electric energy a day. This energy is equivalent to work, since it can be converted into work with almost \(100 \%\) efficiency by an electric motor. (a) If this energy is generated by power plants with an average efficiency of \(0.30,\) how much heat is dumped into the environment each day? (b) How much water would be required to absorb this heat if the water temperature is not to increase more than \(2.0^{\circ} \mathrm{C} ?\)

An oil-burning electric power plant uses steam at \(773 \mathrm{K}\) to drive a turbine, after which the steam is expelled at \(373 \mathrm{K} .\) The engine has an efficiency of \(0.40 .\) What is the theoretical maximum efficiency possible at those temperatures?
Show that in a reversible engine the amount of heat \(Q_{C}\) exhausted to the cold reservoir is related to the net work done \(W_{\text {net }}\) by $$ Q_{\mathrm{C}}=\frac{T_{\mathrm{C}}}{T_{\mathrm{H}}-T_{\mathrm{C}}} W_{\mathrm{net}} $$
The motor that drives a reversible refrigerator produces \(148 \mathrm{W}\) of useful power. The hot and cold temperatures of the heat reservoirs are \(20.0^{\circ} \mathrm{C}\) and \(-5.0^{\circ} \mathrm{C} .\) What is the maximum amount of ice it can produce in \(2.0 \mathrm{h}\) from water that is initially at \(8.0^{\circ} \mathrm{C} ?\)
A container holding \(1.20 \mathrm{kg}\) of water at \(20.0^{\circ} \mathrm{C}\) is placed in a freezer that is kept at \(-20.0^{\circ} \mathrm{C} .\) The water freezes and comes to thermal equilibrium with the interior of the freezer. What is the minimum amount of electrical energy required by the freezer to do this if it operates between reservoirs at temperatures of $20.0^{\circ} \mathrm{C}\( and \)-20.0^{\circ} \mathrm{C} ?$
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