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Section 25.6 Second Law Bootcamp

Subsection 25.6.1 Heat Engine

Subsection 25.6.2 Carnot Engine

Subsection 25.6.3 Carnot Refrigerator

Subsection 25.6.4 Real Engines

Subsection 25.6.5 Miscellaneous

An engine working in a Carnot cycle between two heat baths of temperatures \(600\text{ K}\) and \(273\text{ K}\) runs an electric motor that uses \(60\text{ W}\) of power. If each cycle is completed in 10 seconds, how much heat does the engine absorb from the higher-temperature bath in each cycle?


Use \(W = P\Delta t\) to find the work the engine should produce.


\(1100\ \text{J}\text{.}\)


The efficiency

\begin{equation*} \eta = \frac{T_H - T_C}{T_H} = \frac{327}{600} = 0.545. \end{equation*}

The work output in each cycle is

\begin{equation*} W = P\Delta t = 60\ \text{W}\times 10\ \text{s} = 600\ \text{J}. \end{equation*}


\begin{equation*} Q_{in} = \frac{W}{\eta} = \frac{600\ \text{J}}{0.545} = 1100\ \text{J}. \end{equation*}

A Carnot cycle working between \(100^{\circ}\text{C}\) and \(30^{\circ}\text{C}\) is used to drive a refrigerator between \(-10^{\circ}\text{C}\) and \(30^{\circ}\text{C}\text{.}\) (a) How much energy must the Carnot engine produce per second so that the refrigerator is able to discard 10 Joules of energy per second? (b) How much heat does the Carnot engine take in each second from its hot bath source to produce the work required to run the refrigerator at \(10\text{ J/s}\) rate?


First find the work required to discard 10 J.


\(8.1\text{ J}\)

Solution 1 (a)

(a) The definition of the coefficient of performance of the refrigerator gives the work required to extract \(Q_C\) from the refrigerator.

\begin{equation*} W = \frac{Q}{\beta}. \end{equation*}

The coefficient of performance \(\beta\) for a Carnot refrigerator can be computed from the temperatures inside and outside.

\begin{equation*} \beta = \frac{T_C}{T_H-T_C} = \frac{263.15}{40} = 6.58. \end{equation*}

Therefore we need the engine to supply the following energy per second.

\begin{equation*} W = \frac{Q}{\beta} = \frac{10\ \text{J}}{6.58} = 1.52\ \text{J}. \end{equation*}
Solution 2 (b)

(b) Since we need \(1.52\text{ J}\) of energy from the engine we need

\begin{align*} Q_{in} \amp = \frac{W}{\eta} = \frac{W T'_H}{T'_H - T'_C} \\ \amp = \frac{1.52\ \text{J} \times 373.15}{ 70} = 8.1\ \text{J}. \end{align*}

Thus, in the end we need to spend \(8.1\text{J}\) of thermal energy from the ultimate source in the engine/refrigerator combined system to take out \(10\text{ J}\) of energy from inside the refrigerator assuming the temperature of the bath at \(30^{\circ}\text{C}\) does not change.

The work output of a Carnot engine operating between temperatures \(T_1\) and \(T_2\) with \(T_1>T_2\) is used to drive a refrigerator between temperatures \(T_3\) and \(T_4\) where \(T_3>T_4\text{.}\) Find the ratio of heat taken from thermal baths \(T_1\) and \(T_4 \) in terms of the four temperatures.

Figure 25.6.17. Figure for Problem 25.6.16.

Work in one cycle of engine will be the work used by the refrigerator.


\(\left(T_3/T_4 -1 \right)/\left(1-T_2/T_1 \right)\text{.}\)


In this problem, we have a Carnot engine and a Carnot refrigrator. Let us denote the quantities of refrigerator by a prime and those of the engine without a prime. We wish the work produced by the engine to be equal to the work used by the refrigerator.

\begin{equation*} Q_H - Q_C = Q^\prime_H - Q^\prime_C. \end{equation*}

Dividing both sides by $Q_H Q'_C$ gives

\begin{equation*} \frac{1}{Q^\prime_C} \left(1 - \frac{Q_C}{Q_H}\right) = \frac{1}{Q_H} \left( \frac{Q^\prime_H}{Q^\prime_C} - 1\right). \end{equation*}

Carnot cycles have

\begin{equation*} \frac{Q_H}{Q_C} = \frac{T_1}{T_2}\ \ \textrm{and}\ \ \frac{Q^\prime_H}{Q^\prime_C} = \frac{T_3}{T_4}. \end{equation*}


\begin{equation*} \frac{Q_H}{Q^\prime_C} = \frac{\frac{T_3}{T_4} - 1}{1 - \frac{T_2}{T_1}} = \left( \frac{T_3-T_4}{T_1-T_2}\right)\ \frac{T_1}{T_4}. \end{equation*}