The Workbook

Kelvin-Planck: no engine working in a cycle can take heat from a single reservoir and turn all of it into work. This is about engines.

Clausius: no device working in a cycle can move heat from a colder body to a hotter body with no other effect. This is about refrigerators.

\(\eta = \dfrac{W}{Q_1} = \dfrac{Q_1 - Q_2}{Q_1} = 1 - \dfrac{Q_2}{Q_1}\).

\(W = 1000 - 600 = 400\) J. \(\eta = \dfrac{400}{1000} = 0.4 = 40\%\).

\(\eta = 1\) would need \(Q_2 = 0\): all the heat taken from the source would become work, with nothing given to a sink. The Kelvin-Planck statement forbids exactly this. Some heat must always be rejected to a colder body.

1. \(AB\): isothermal expansion at \(T_1\). \(BC\): adiabatic expansion, \(T_1\) to \(T_2\). \(CD\): isothermal compression at \(T_2\). \(DA\): adiabatic compression, \(T_2\) back to \(T_1\).

2. Heat \(Q_1\) is absorbed in \(AB\) (the gas expands at constant \(T\), so it must take in heat). Heat \(Q_2\) is rejected in \(CD\). No heat flows in \(BC\) or \(DA\).

3. The enclosed area is the net work done per cycle, \(W = Q_1 - Q_2\).

\(\eta = 1 - \dfrac{T_2}{T_1} = 1 - \dfrac{300}{500} = 0.4 = 40\%\).

Sink at \(250\) K: \(\eta = 1 - \dfrac{250}{500} = 0.5 = 50\%\).

\(\eta = 0.5\) with \(T_2 = 300\) K: \(\dfrac{300}{T_1} = 0.5\), so \(T_1 = 600\) K. The source must be raised by \(100\) K, while the sink needed to fall by only \(50\) K for the same gain.

\(\eta = 1 - \dfrac{300}{400} = 0.25\). \(W = \eta Q_1 = 250\) J. \(Q_2 = Q_1 - W = 750\) J.

\(\dfrac{Q_2}{Q_1} = \dfrac{750}{1000} = 0.75\) and \(\dfrac{T_2}{T_1} = \dfrac{300}{400} = 0.75\). For a Carnot engine the heats are in the ratio of the absolute temperatures.

No engine working between two given temperatures can be more efficient than a Carnot engine between the same temperatures. All Carnot engines between those temperatures have the same efficiency.

So \(\eta\) depends only on \(T_1\) and \(T_2\), not on the gas or any other working substance.

1. \(Q_2 = Q_1 - W = 800 - 200 = 600\) J.

2. \(\eta = \dfrac{W}{Q_1} = \dfrac{200}{800} = 0.25 = 25\%\).

3. No engine can beat Carnot, so \(1 - \dfrac{300}{T_1} \geq 0.25\), giving \(T_1 \geq 400\) K. The lowest possible source temperature is \(400\) K.

1. Energy is conserved: \(Q_1 = Q_2 + W = 300 + 100 = 400\) J.

2. \(\beta = \dfrac{\text{heat removed from the cold space}}{\text{work supplied}} = \dfrac{Q_2}{W} = \dfrac{300}{100} = 3\).

3. Heat does go from cold to hot, but not by itself: work is supplied, and the room receives more heat than the inside loses. The Clausius statement forbids only the transfer with no other effect.

\(\beta = \dfrac{Q_2}{Q_1 - Q_2} = \dfrac{T_2}{T_1 - T_2}\) for a Carnot cycle run backwards.

\(\beta = \dfrac{270}{300 - 270} = 9\). Each joule of work moves \(9\) J of heat out of the refrigerator.

\(\beta_{hp} = \dfrac{Q_1}{W} = \dfrac{Q_1}{Q_1 - Q_2} = \dfrac{T_1}{T_1 - T_2}\).

\(\beta_{hp} = \dfrac{300}{30} = 10\). Since \(Q_1 = Q_2 + W\), \(\dfrac{Q_1}{W} = \dfrac{Q_2}{W} + 1\), so \(\beta_{hp} = \beta + 1 = 9 + 1 = 10\). A heat pump delivers \(10\) J of heat to the house for each joule of electrical work.

\(\eta = 1 - \dfrac{T_2}{T_1}\), so \(\dfrac{T_2}{T_1} = 1 - \eta\). \(\beta = \dfrac{T_2}{T_1 - T_2} = \dfrac{T_2/T_1}{1 - T_2/T_1} = \dfrac{1 - \eta}{\eta}\).

\(\eta = 0.4\): \(\beta = \dfrac{0.6}{0.4} = 1.5\).

Warmer. The refrigerator takes heat \(Q_2\) from the air in front of it and gives \(Q_1 = Q_2 + W\) to the air behind it. The net effect on the room is the work \(W\), which all ends up as heat. The room gains energy at the rate of the refrigerator's power.

1. \(T_2 = 300\) K: \(\eta = 1 - \dfrac{300}{600} = 0.5\). \(T_2 = 450\) K: \(\eta = 1 - \dfrac{450}{600} = 0.25\).

2. \(\eta = 1\) needs \(Q_2 = 0\), that is \(T_2 = 0\) K. A sink at absolute zero cannot be reached, so \(\eta\) is always less than \(1\).

3. \(\eta = 0\) when \(T_2 = T_1 = 600\) K. With both reservoirs at the same temperature the engine does no work: all the heat taken in is given straight back.

\(1 - \dfrac{T_2}{T_1} = \dfrac{1}{6}\) gives \(T_2 = \dfrac{5}{6}T_1\). \(1 - \dfrac{T_2 - 62}{T_1} = \dfrac{1}{3}\) gives \(T_2 - 62 = \dfrac{2}{3}T_1\).

Subtract: \(62 = \dfrac{5}{6}T_1 - \dfrac{2}{3}T_1 = \dfrac{1}{6}T_1\), so \(T_1 = 372\) K and \(T_2 = 310\) K.

1. Engine \(1\): \(W_1 = Q_1\left(1 - \dfrac{T}{800}\right)\) and \(Q = Q_1\dfrac{T}{800}\). Engine \(2\): \(W_2 = Q\left(1 - \dfrac{200}{T}\right) = Q_1\dfrac{T - 200}{800}\).

   \(W_1 = W_2\): \(800 - T = T - 200\), so \(T = 500\) K, the mean of the two temperatures.

2. \(1 - \dfrac{T}{800} = 1 - \dfrac{200}{T}\), so \(T^2 = 800 \times 200\) and \(T = 400\) K, the geometric mean.

\(\beta = \dfrac{T_2}{T_1 - T_2} = \dfrac{250}{50} = 5\). Power \(= \dfrac{\text{heat removed per second}}{\beta} = \dfrac{600}{5} = 120\) W.

No. The Carnot efficiency between \(500\) K and \(300\) K is \(1 - \dfrac{300}{500} = 40\%\), and no engine between these temperatures can do better. A \(60\%\) engine would break the second law.

In the isothermal expansion \(\Delta U = 0\), so the heat absorbed equals the work done: \(Q_1 = 2301\) J.

\(\eta = 1 - \dfrac{300}{400} = 0.25\). \(W = \eta Q_1 = 575\) J. \(Q_2 = Q_1 - W = 1726\) J.

Hot body: \(\Delta S = \dfrac{-300}{300} = -1\) J/K. Cold body: \(\Delta S = \dfrac{+300}{200} = +1.5\) J/K. Total: \(+0.5\) J/K.

The total entropy rises. This is the mark of an irreversible process: heat flowing from hot to cold happens by itself and never reverses on its own.

A perpetual motion machine of the second kind is an engine with no sink. It takes heat from a single reservoir, such as the sea, and turns all of it into work, cycle after cycle. It would not break the first law, since energy is conserved. It breaks the Kelvin-Planck statement of the second law, so it cannot be built.

\(dS = \dfrac{dQ_{rev}}{T}\), the heat absorbed reversibly divided by the absolute temperature at which it is absorbed. Unit: J/K.

Entropy is a state function. Its change between two states is the same whatever the path, though it must be calculated along a reversible path.

\(\Delta S = \dfrac{Q}{T} = \dfrac{1000}{500} = +2\) J/K.

The reservoir gives out \(1000\) J at \(500\) K, so its entropy change is \(-2\) J/K. For this reversible exchange the total is zero.

Melting happens at constant temperature: \(\Delta S = \dfrac{mL}{T} = \dfrac{3.36 \times 10^5}{273} = 1231\) J/K.

\(\Delta S = \dfrac{mL}{T} = \dfrac{2.26 \times 10^6}{373} = 6059\) J/K, about \(6.06\) kJ/K.

Boiling needs nearly seven times the latent heat of melting, at a temperature only \(37\%\) higher. Turning a liquid into a widely spread gas is a far larger increase in disorder than loosening a solid into a liquid.

Each small step absorbs \(dQ = ms\,dT\) at temperature \(T\): \(dS = \dfrac{ms\,dT}{T}\). Integrating from \(T_1\) to \(T_2\): \(\Delta S = ms\ln\dfrac{T_2}{T_1}\).

\(\Delta S = 1 \times 400 \times \ln 2 = 400 \times 0.693 = 277\) J/K.

Isothermal: \(\Delta U = 0\), so \(dQ = dW = P\,dV = \dfrac{nRT}{V}dV\). Then \(dS = \dfrac{dQ}{T} = nR\dfrac{dV}{V}\), and integrating gives \(\Delta S = nR\ln\dfrac{V_2}{V_1}\).

\(\Delta S = 1 \times 8.3 \times 0.693 = 5.75\) J/K.

Constant volume: \(\Delta S = nC_V\ln\dfrac{T_2}{T_1} = 12.45 \times 0.693 = 8.63\) J/K.

Constant pressure: \(\Delta S = nC_P\ln\dfrac{T_2}{T_1} = 20.75 \times 0.693 = 14.4\) J/K.

At constant pressure the gas also expands, and more heat must be absorbed for the same rise in temperature. More heat absorbed at the same temperatures means a larger entropy change.

\(dQ = 0\) at every step, so \(dS = \dfrac{dQ_{rev}}{T} = 0\) and \(\Delta S = 0\). The process is isentropic: it is a vertical line on a \(T\)-\(S\) diagram.

1. \(ab\): isothermal expansion at \(500\) K. \(bc\): adiabatic expansion. \(cd\): isothermal compression at \(300\) K. \(da\): adiabatic compression. An isothermal process has constant \(T\), so it is horizontal. A reversible adiabatic process has constant \(S\), so it is vertical.

2. Heat is the area under the line on a \(T\)-\(S\) diagram. \(ab\): \(Q_1 = T_1\Delta S = 500 \times 2 = 1000\) J absorbed. \(cd\): \(Q_2 = 300 \times 2 = 600\) J rejected.

3. \(W = Q_1 - Q_2 = 400\) J, the area of the rectangle. \(\eta = \dfrac{400}{1000} = 40\%\), and \(1 - \dfrac{300}{500} = 0.4\) agrees.

1. \(T\) is constant along \(PQ\), so it is an isothermal process. The entropy rises, so heat is absorbed.

2. The area under a \(T\)-\(S\) curve is \(\int T\,dS = Q\), the heat absorbed. \(Q = 400 \times (3 - 1) = 800\) J.

3. For an ideal gas \(\Delta U = 0\) at constant \(T\), so \(W = Q = 800\) J.

\(AB\) is the isothermal expansion at the higher temperature \(T_1\). Heat is absorbed, so \(S\) rises: it is \(ab\), the upper horizontal line going right.

\(BC\) is the adiabatic expansion: \(S\) is constant and \(T\) falls, so it is \(bc\), the right vertical line going down.

\(CD\) is the isothermal compression at \(T_2\), heat rejected, \(S\) falls: it is \(cd\), the lower line going left. \(DA\) is the adiabatic compression, \(T\) rising at constant \(S\): it is \(da\).

1. \(XY\): area under the sloping line, a trapezium, \(\dfrac{300 + 500}{2} \times 2 = 800\) J absorbed. \(ZX\): \(300 \times 2 = 600\) J rejected. No heat flows along \(YZ\).

2. \(W = 800 - 600 = 200\) J, the area of the triangle. \(\eta = \dfrac{200}{800} = 25\%\).

3. \(YZ\) is vertical: \(S\) is constant, so it is a reversible adiabatic (isentropic) process.

Entropy is a state function. After a complete cycle the system is back in its initial state, so \(\Delta S = 0\).

In the Carnot cycle the only entropy changes are \(+\dfrac{Q_1}{T_1}\) in the isothermal expansion and \(-\dfrac{Q_2}{T_2}\) in the isothermal compression. Their sum must be zero, so \(\dfrac{Q_1}{T_1} = \dfrac{Q_2}{T_2}\).

Hot reservoir: \(\dfrac{-1000}{500} = -2\) J/K. Cold reservoir: \(\dfrac{+1000}{250} = +4\) J/K. Universe: \(+2\) J/K.

The entropy of the universe has risen, so the process is irreversible. The heat will never flow back by itself.

No. \(\Delta S\) depends only on the end states, which are the same as for an isothermal expansion to twice the volume. So \(\Delta S = nR\ln 2 = 5.75\) J/K.

\(\Delta S\) must be found along a reversible path between the same states, here the isothermal one. The free expansion itself has \(Q = 0\), but it is irreversible, so \(Q/T\) along it does not give the entropy change.

The entropy rises. Each gas expands into the whole vessel, like a free expansion, and gains \(nR\ln 2\) if the volumes are equal. The mixed state is more disordered and the gases never separate by themselves.

If both sides hold the same gas at the same \(T\) and \(P\), removing the partition changes nothing that can be measured. The entropy does not change.

The entropy of an isolated system, and so of the universe, never decreases: \(\Delta S_{universe} \geq 0\).

In a reversible process \(\Delta S_{universe} = 0\). In an irreversible process, which includes every real process, \(\Delta S_{universe} > 0\).

1. From the graph, about \(14\) J/K. \(C_P\ln 2 = 20.75 \times 0.693 = 14.4\) J/K.

2. About \(23\) J/K: \(C_P\ln 3 = 20.75 \times 1.099 = 22.8\) J/K.

3. \(dS = \dfrac{C_P\,dT}{T}\): the same heat gives a smaller entropy change when the gas is already hot. The slope of the curve is \(\dfrac{C_P}{T}\), which falls as \(T\) rises.

Body: \(\dfrac{-600}{400} = -1.5\) J/K. Surroundings: \(\dfrac{+600}{300} = +2\) J/K. Whole: \(+0.5\) J/K. The body loses entropy but the surroundings gain more.

The cold body loses \(Q_2\) at \(T_2\): \(\Delta S_2 = -\dfrac{Q_2}{T_2}\). The hot body gains \(Q_1\) at \(T_1\): \(\Delta S_1 = +\dfrac{Q_1}{T_1}\). The refrigerator itself returns to its state each cycle.

\(\Delta S_{universe} = \dfrac{Q_1}{T_1} - \dfrac{Q_2}{T_2} \geq 0\), so \(\dfrac{Q_1}{T_1} \geq \dfrac{Q_2}{T_2}\).

If \(W = 0\) then \(Q_1 = Q_2\), and the inequality would need \(T_1 \leq T_2\), which is false. So \(W > 0\): heat cannot pass from cold to hot without work. This is the Clausius statement.

\(\Delta S = ms\ln\dfrac{T_2}{T_1} = 2 \times 4200 \times \ln\dfrac{273}{373} = -8400 \times 0.312 = -2621\) J/K.

The water is not isolated. The heat it loses goes to the cooler surroundings. Their entropy rises by more than \(2621\) J/K, because they receive the heat at a lower temperature. The total still increases.