Second Law of Thermodynamics

(b) 47%

The maximum (Carnot) efficiency uses kelvin temperatures: $T_1 = 300 + 273 = 573$ K and $T_2 = 30 + 273 = 303$ K.

If a small amount of heat $dQ$ is given to (or taken from) a system reversibly at absolute temperature $T$, the change in its entropy is

For a reversible change at constant temperature, $\Delta S = \dfrac{Q}{T}$. The SI unit of entropy is J K$^{-1}$.

(a) Two isothermal and two adiabatic processes.

A Carnot cycle has four reversible steps: isothermal expansion at $T_1$, adiabatic expansion (cooling from $T_1$ to $T_2$), isothermal compression at $T_2$, and adiabatic compression (back to $T_1$).

Given:

• Source temperature, $T_1 = 600$ K
• Sink temperature, $T_2 = 300$ K
• Work output per cycle, $W = 800$ J
• Heat supplied per cycle, $Q_1 = ?$

Efficiency of the Carnot engine:

Also $\eta = \dfrac{W}{Q_1}$, so

(Heat rejected to the sink: $Q_2 = Q_1 - W = 800$ J.)

(a) 400 J

$\eta = 1 - \dfrac{T_2}{T_1} = 1 - \dfrac{300}{600} = 0.5$. Since $\eta = W/Q_1$, the energy supplied is $Q_1 = \dfrac{W}{\eta} = \dfrac{200}{0.5} = 400$ J.

Entropy is a thermodynamic property of a system that measures its degree of disorder (randomness) of the molecules. It also measures how much of the system's thermal energy is unavailable for doing work. When heat $dQ$ is added reversibly at absolute temperature $T$, the change in entropy is

SI unit: J K$^{-1}$.

First law in terms of entropy: the first law is $dQ = dU + P\,dV$. For a reversible process $dQ = T\,dS$, so

No. A refrigerator takes heat $Q_2$ from its inside (here, from the room through the open door). Using electrical work $W$, it gives out heat $Q_1 = Q_2 + W$ through the coils at its back, which is also into the room. So the room receives more heat than it loses: there is a net gain of $W$ in every cycle, and the room actually becomes slightly warmer, not cooler.

(B) 50%

Taking it as an ideal (Carnot) engine: $T_1 = 327 + 273 = 600$ K and $T_2 = 27 + 273 = 300$ K.

(A) 26.8%

Steam point $T_1 = 373$ K and ice point $T_2 = 273$ K.

Given:

• Initial efficiency, $\eta_1 = 25\% = 0.25$
• Sink temperature, $T_2 = 9^\circ$C $= 9 + 273 = 282$ K (kept the same)
• New efficiency, $\eta_2 = 50\% = 0.50$

For a Carnot engine, $\eta = 1 - \dfrac{T_2}{T_1}$, so $T_1 = \dfrac{T_2}{1 - \eta}$.

Initial source temperature:

New source temperature:

Increase required:

The source temperature must be raised by 188 degrees.

(c) $\eta_1 < \eta_2$

A diesel engine works at a much higher compression ratio (about 15–20) than a petrol engine (about 6–10). The efficiency of an internal combustion engine increases with compression ratio ($\eta = 1 - 1/\rho^{\gamma-1}$ for the petrol cycle), so the diesel engine is more efficient.

Given:

• Petrol used, $m = 10$ kg $= 10^4$ g per hour, i.e. $t = 3600$ s
• Calorific value, $c = 1.14\times10^4$ cal g$^{-1}$
• Output power, $P_{\text{out}} = 30$ kW $= 3.0\times10^4$ W
• $J = 4.2$ J cal$^{-1}$

Heat energy supplied per hour:

Input power:

Efficiency:

(Using $J = 4.186$ J cal$^{-1}$ gives practically the same result, about 22.6%.)

(b) States that heat cannot spontaneously flow from a colder object to a hotter object.

The Clausius statement says that heat cannot by itself pass from a colder body to a hotter one without external work (as in a refrigerator). Its significance is that it fixes the natural direction of heat flow.

(a) First law: The heat supplied to a system equals the increase in its internal energy plus the work done by the system: $Q = \Delta U + W$.

In an adiabatic process $Q = 0$, so $\Delta U = -W$. When the gas expands, $W>0$, so $\Delta U<0$. The gas does work at the expense of its own internal energy, so its temperature falls and cooling is observed.

(b) For the Carnot engine, $Q_1/T_1 = Q_2/T_2$:

$Q_2 = 1000\times\frac{300}{600} = 500\ \text{J}$

The work output is $W = Q_1 - Q_2 = 500$ J, and this is the work input of the refrigerator.

For the Carnot refrigerator, $Q_3/Q_4 = T_{3}/T_{4} = 500/400 = 1.25$, and $Q_3 = Q_4 + W$:

$1.25Q_4 = Q_4 + 500 \;\Rightarrow\; Q_4 = 2000\ \text{J}, \qquad Q_3 = 2500\ \text{J}$

So $Q_2 = 500$ J, $Q_3 = 2500$ J and $Q_4 = 2000$ J.

(a) P–V diagram of a petrol engine (Otto cycle) and its working

   P
   ^
   |    C
   |    *
   |    |*
   |    | *            C -> D  adiabatic expansion (power stroke)
   |    |  *
   | Q1 |   **
   | in |     **
   |    |       ***
   |    B          ****
   |     *             *****
   |      **                ****** D
   |        ***                    |
   |           ****                |  D -> A  constant volume,
   |  A -> B       *****           |          heat Q2 rejected
   |  adiabatic         ******     |
   |  compression             *****|
   |    E--------------------------A   E -> A suction, A -> E exhaust
   |                                   (atmospheric pressure)
   +----+--------------------------+--------> V
        V2                         V1

   B -> C : spark ignites the mixture, heat Q1 added at constant volume

A four-stroke petrol engine works on the Otto cycle. Its strokes are:

1. Suction stroke (E → A): the inlet valve opens and the piston moves out. A mixture of petrol vapour and air is drawn into the cylinder at about atmospheric pressure, and the volume increases to $V_1$.
2. Compression stroke (A → B): both valves are closed. The piston moves in and compresses the mixture adiabatically to volume $V_2$. Its pressure and temperature rise.
3. Ignition (B → C): at the end of compression a spark from the spark plug ignites the mixture. It burns almost instantly, so heat $Q_1$ is added at constant volume, and the pressure and temperature rise sharply.
4. Power (working) stroke (C → D): the hot gases expand adiabatically and push the piston out. This is the only stroke in which useful work is done.
5. Exhaust (D → A → E): the exhaust valve opens. The pressure falls at constant volume (D → A), rejecting heat $Q_2$. The piston then moves in and pushes the burnt gases out (A → E). The cycle repeats.

The net work per cycle is the area enclosed by the loop ABCDA. The efficiency is

(b) Petrol engine and diesel engine compared

Since efficiency increases with compression ratio, and a diesel engine works at a much higher compression ratio, a diesel engine is more efficient than a petrol engine.

(d) 10

For an ideal refrigerator, $\beta = \dfrac{T_2}{T_1 - T_2}$. With $T_2 = 273$ K (ice point) and $T_1 = 300$ K (27°C):

(i) Second law of thermodynamics

• Kelvin–Planck statement: it is impossible to construct an engine which, working in a cycle, absorbs heat from a source and converts all of it into work without rejecting some heat to a colder sink.
• Clausius statement: it is impossible for a self-acting machine, unaided by any external agency, to transfer heat from a colder body to a hotter body.

(ii) Does a refrigerator violate the second law?

No. The Clausius statement only says that heat cannot flow from cold to hot by itself. A refrigerator does not do this by itself. Its compressor, run by electrical energy, does external work $W$ on the refrigerant. Using this work, it takes heat $Q_2$ from the cold inside and gives out heat $Q_1 = Q_2 + W$ to the warmer room. Since an external agency (the work input) is needed, the second law is obeyed.

(iii) (a) Efficiency of the heat engine

        +-------------------+
        |  Source  T1       |
        +---------+---------+
                  |  Q1
                  v
             +---------+
             | Engine  |=======>  W = Q1 - Q2
             +---------+
                  |  Q2
                  v
        +---------+---------+
        |  Sink  T2         |
        +-------------------+

The efficiency is the fraction of the heat absorbed that is converted into work. By the first law, for one cycle $W = Q_1 - Q_2$, so

For an ideal (Carnot) engine, $\dfrac{Q_2}{Q_1} = \dfrac{T_2}{T_1}$, so $\eta = 1 - \dfrac{T_2}{T_1}$.

(iii) (b) When is the efficiency zero?

$\eta = 0$ when $Q_2 = Q_1$, i.e. when all the heat taken in is rejected and no work is done ($W = 0$). For an ideal engine, $\eta = 1 - T_2/T_1 = 0$ when $T_1 = T_2$, i.e. when the source and sink are at the same temperature. With no temperature difference, no heat flows through the engine and no work can be obtained. (At the other limit, $\eta = 100\%$ would need $T_2 = 0$ K, which is impossible.)