First Law of Thermodynamics and Heat Engines

Beginner

Rub your hands together and they warm up; pump up a bicycle tyre and the pump gets hot. In both cases work has turned into heat.

Step 1: three words. Temperature tells you how fast, on average, the molecules of something jiggle. Thermal energy is the total energy of all those molecules. Heat is energy on the move, flowing from a hotter object to a colder one because their temperatures differ.

Step 2: the first law. Energy is never created or destroyed, only moved or changed in form. Heat a gas in a cylinder and the energy you put in does one of two things, or a bit of both: it warms the gas, or it lets the gas push a piston, which is work.

Step 3: heat engines. A car engine or a power station lets heat flow from a hot place to a cold place and catches part of it as work. Some heat must always flow on to the cold side as waste: that is why cars have radiators and power stations have cooling towers.

Step 4: the bigger the gap between the hot side and the cold side, the larger the share of the heat that can become work.

Check yourself: you give a gas 100 J of heat and it does 30 J of work pushing a piston. Where are the other 70 J? (In the gas, which is now warmer.)

In the animation, a gas in a cylinder repeats four steps: heated, it pushes the piston up and cools as it keeps expanding; then it is squeezed back while it hands waste heat to the cold side. Squeezing the cooled gas costs less work than the hot gas gave out: the difference is the useful work.

Intermediate

Heat, work and internal energy

A gas can take in heat and grow no warmer, if it spends that energy pushing a piston. Thermodynamics follows energy into and out of a system, here the gas in the cylinder, with three quantities measured in joules. The internal energy $U$ is the energy stored in the gas; for an ideal gas it is the kinetic energy of its molecules. Heat $Q$ crosses the boundary because of a temperature difference, and work $W$ crosses it when a force moves something, as when the gas pushes a piston.

$Q$ is positive when heat flows into the gas. $W$ is the work done by the gas: positive when it expands, negative when the piston squeezes it. At constant pressure $W=p\Delta V$, so a gas at 100 kPa that expands by 2 L does $100\mathrm{kPa}\cdot 2\mathrm{L}=200\mathrm{J}$ of work, since a kilopascal times a litre is a joule.

The first law: where the energy goes

Heat is energy on the move, so the energy that enters as heat either stays in the gas or leaves as work:
$\Delta U=Q-W$

This is the first law of thermodynamics. In the 1840s Joule warmed water with paddles driven by falling weights and found that about 4.2 J of work does what one calorie of heat does.

For an ideal monatomic gas $U=\frac{3}{2}nRT$, so $\Delta U=\frac{3}{2}nR\Delta T$. The animation's gas has $nR=2J/K$ (about 0.24 mol), so its internal energy changes by 3 J per kelvin. Four simple processes share it out differently:

• Isothermal, at constant $T$: $\Delta U=0$, so $Q=W$. In the animation's first stroke the gas absorbs 1099 J and does 1099 J of work.
• Adiabatic, with no heat exchanged: $Q=0$, so $W=-\Delta U$. In the second stroke the gas does 600 J of work and cools from 500 K to 300 K; compressed, a gas heats up, as in a bicycle pump.
• Constant volume: the piston cannot move, so $W=0$ and all the heat raises the temperature, $Q=\Delta U$.
• Constant pressure: $W=p\Delta V=nR\Delta T$. Heating 1 mol of a monatomic gas by 10 K takes $\Delta U=125\mathrm{J}$ plus $W=83\mathrm{J}$, so $Q=208\mathrm{J}$.

Heat engines and efficiency

A heat engine repeats a cycle, bringing the gas back to its starting state, so over each cycle $\Delta U=0$. It takes heat $Q_h$ from a hot reservoir and gives heat $Q_c$ to a cold one, both counted as positive amounts. Its net work, and its efficiency (the fraction of the heat taken in that becomes work), are

• $W=Q_h-Q_c$
• $\eta =\frac{W}{Q_h}=1-\frac{Q_c}{Q_h}$

A reservoir is a body too large for its temperature to change, such as a furnace or a river. Between reservoirs at absolute temperatures $T_h$ and $T_c$, no engine can beat the Carnot efficiency (the expert level shows why):
${\eta }_C=1-\frac{T_c}{T_h}$
Some heat must always be rejected; the bigger the temperature gap, the larger the share that can become work.

Worked example: the engine in the animation

The animated engine runs a Carnot cycle between $T_h=500\mathrm{K}$ and $T_c=300\mathrm{K}$. In stroke 1 the gas expands isothermally from 1 L to 3 L on the hot reservoir. An ideal gas at constant temperature does work $nRT\ln (V_f/V_i)$ (derived at the expert level), all of it supplied as heat:
$Q_h=2\cdot 500\cdot \ln 3=1098.6\mathrm{J}$

In stroke 2 the insulated gas expands on to 6.46 L, doing $W=-\Delta U=3\cdot 200=600\mathrm{J}$ of work as it cools to 300 K. In stroke 3 it is squeezed on the cold reservoir from 6.46 L to 2.15 L, again by a factor of 3, and rejects $Q_c=2\cdot 300\cdot \ln 3=659.2\mathrm{J}$. Stroke 4, adiabatic, costs 600 J and restores 1 L and 500 K. Then
$W=Q_h-Q_c=1098.6-659.2=439.4\mathrm{J},\eta =\frac{439.4}{1098.6}=40\%$

That is exactly $1-300/500$: a Carnot engine reaches the limit. The two adiabatic strokes cancel, $+600$ and $-600\mathrm{J}$, because each spans the same 200 K.

Reading the cycle: readouts and p–V diagram

The readouts are running totals since the cycle began, rounded to the joule; $\Delta U=Q-W$ holds at every instant. In stroke 1 the red $Q_h$ arrow is on, $T$ stays at 500 K, and $Q$ and $W$ climb together to 1099 J. In stroke 2, $Q$ is frozen while $W$ climbs to 1699 J and $\Delta U$ falls to $-600\mathrm{J}$.

The last two strokes compress the gas. In stroke 3 the blue $Q_c$ arrow appears, and $Q$ and $W$ fall together, to 439 J and 1039 J. In stroke 4, $Q$ stays at 439 J while $W$ falls to 439 J and $\Delta U$ returns to zero: the net heat absorbed has all become work. The green $W$ arrow points up while the gas pushes the piston and down while the piston pushes the gas.

On a pressure–volume diagram the work of a stroke is the area under its curve. Going out along the top of the loop the gas does 1698.6 J; coming back along the bottom the piston does 1259.2 J on it, so the loop encloses the net 439.4 J.

Worked example: a power station

A power station raises steam at 527 °C (800 K), condenses it with river water at 27 °C (300 K), and turns 40 % of its heat into 600 MW of electric power. The cycle equations hold per second too, in megawatts.

The Carnot limit is $1-300/800=62.5\%$, so 40 % is allowed. The fuel must supply $Q_h=W/\eta =600/0.40=1500\mathrm{MW}$, and $Q_c=1500-600=900\mathrm{MW}$ leaves as waste heat, more than the electricity. Water has a specific heat of 4186 J/(kg·K), so warming the river by only 10 K takes $900\times 10^6/(4186\cdot 10)\approx 21500\mathrm{kg}$ of water every second. A perfect Carnot engine would still need $600/0.625=960\mathrm{MW}$ of heat and reject 360 MW.

Common mistakes

• Heat is not stored in a body; it is energy in transit, so $Q$ and $W$ depend on the path. From 1 L at 1000 kPa to 3 L at 333 kPa the gas does 2000 J of work if it first expands at constant pressure, only 667 J if it first cools at constant volume, with $\Delta U=0$ both times.
• Keep one sign convention. Chemists often count the work done on the gas and write $\Delta U=Q+W$; either way, 250 J of heat in and 90 J of work out give $\Delta U=160\mathrm{J}$.
• Adiabatic does not mean isothermal: with no heat exchanged, the gas of stroke 2 cools by 200 K. Nor does isothermal mean no heat: in stroke 1 the gas absorbs 1099 J.
• Put temperatures in kelvins. Between 100 °C and 20 °C the limit is $1-293/373=21\%$, not $1-20/100=80\%$.
• Waste heat is not a design flaw: even a frictionless engine between 500 K and 300 K must reject at least 60 % of its heat.

Where you meet it

Car engines turn roughly a quarter to a third of their fuel's energy into work; steam power stations reach about 40 %, and combined-cycle gas plants about 60 % by starting from much hotter gas. Refrigerators and heat pumps are heat engines run backwards. Fuel cells and batteries make electrical work without passing through heat, so the Carnot limit does not bind them. The next lesson, The Second Law of Thermodynamics and Entropy, explains why every engine must reject heat and turns the ratio $Q/T$ into entropy.

Advanced

Heat and work depend on the path

An engine turns heat into work by taking a gas round a closed loop of states again and again. Before asking how efficient that can be, we need to see why a loop can deliver any work at all. The reason is that heat and work, unlike energy, depend on the route the gas takes from one state to another.

The gas throughout is the animation's: a monatomic ideal gas, about 0.24 mol of it, so that $nR=2.00J/K$, where $n$ is the amount of gas and $R$ the gas constant. Its state is fixed by its pressure $p$ and volume $V$, and then its temperature follows from $pV=nRT$. Its internal energy $U$, the total kinetic energy of its atoms, is $U=\frac{3}{2}nRT$. Write $Q$ for heat flowing into the gas and $W$ for work done by the gas.

A change is quasi-static when it is slow enough for the gas to stay uniform, so that $p$ and $T$ have definite values at every moment. For one small quasi-static step the first law, conservation of energy, reads
$dU=\delta Q-\delta W,\delta W=pdV$
The second formula is force times distance: the gas pushes a piston of area $A$ with force $pA$, and if the piston moves out by $dx$ the gas does work $pAdx=pdV$. It is positive when the gas expands and negative when it is compressed.

The internal energy is a function of the state: once $p$ and $V$ are known, so is $U$ (for an ideal gas, $T$ alone fixes it). Its change $dU$ is therefore an exact differential, and round any closed loop $\oint dU=0$. Heat and work are different. The symbol $\delta$ marks a small amount whose total depends on the path, so a phrase such as "the heat in the gas" has no meaning. Figure 1 shows this directly.

Legend

$p$	Pressure of the gas.	$\mathrm{Pa}$
$V$	Volume of the gas.	${\mathrm{m}}^3$
$a$	Route along the 500 K isotherm, at constant temperature.
$b$	Route that first expands at constant pressure, then cools at constant volume.
$c$	Route that first cools at constant volume, then expands at constant pressure.
$W_a$	Work done by the gas along route $a$: the area under it.	$\mathrm{J}$
$W_b$	Work done by the gas along route $b$: the area under it.	$\mathrm{J}$
$W_c$	Work done by the gas along route $c$: the area under it.	$\mathrm{J}$

Figure 1 draws three routes the gas can take from state 1 (1 L, 1000 kPa) to state 2 (3 L, 333 kPa). In each panel $p$ runs up and $V$ across. Look at the shaded area under each route. Every small step adds $pdV$ to the work, a thin strip of height $p$ and width $dV$, so the work of a route is the area under it. A kilopascal times a litre is $10^3\mathrm{Pa}\cdot 10^{-3}{\mathrm{m}}^3=1\mathrm{J}$, so areas in these units read directly in joules.

Route $a$ follows the isotherm $pV=nRT=2.00\cdot 500=1000\mathrm{J}$, the curve of constant temperature 500 K through both states. Its area, worked out in the third section, is $W_a=1098.6\mathrm{J}$. Route $b$ first expands at 1000 kPa and then cools at constant volume; the cooling step is vertical and adds no area, so the area under route $b$ is a rectangle of height 1000 kPa and width 2 L: $W_b=2000\mathrm{J}$. Route $c$ first cools at constant volume and then expands at 333 kPa, which leaves a rectangle of height 333 kPa: $W_c=667\mathrm{J}$.

Both states lie on the 500 K isotherm, so $\Delta U=0$ by every route, and the first law gives $Q=W$: the gas takes in 1098.6 J, 2000 J or 667 J of heat depending on the route. The end states are the same in all three panels, yet the work and the heat differ. They belong to the route, not to the states.

A cycle exploits this freedom. After a full loop the gas is back where it started, so $\oint dU=0$ and $\oint \delta Q=\oint \delta W$. Because the work depends on the path, going out by one route and back by another leaves a net work, the area enclosed by the loop:
$W=\oint pdV$
That net work is paid for by a net intake of heat. This is how every heat engine works.

The Carnot loop on the p–V diagram

The Carnot cycle is the loop that exchanges heat at only two fixed temperatures. It sets the upper limit for every engine, so it is worth reading closely. Its heat comes from two reservoirs, bodies so large that their temperatures do not change when they give or take heat: a hot one at $T_h=500\mathrm{K}$ and a cold one at $T_c=300\mathrm{K}$. The loop is made of two kinds of stroke. On an isotherm the gas touches a reservoir and keeps its temperature. On an adiabat the gas is insulated, so no heat flows at all.

Legend

$p$	Pressure of the gas.	$\mathrm{Pa}$
$V$	Volume of the gas.	${\mathrm{m}}^3$
$V_1$	Volume at corner 1, where the hot isotherm starts.	${\mathrm{m}}^3$
$V_2$	Volume at corner 2, where the hot isotherm ends.	${\mathrm{m}}^3$
$V_3$	Volume at corner 3, where the cold isotherm starts.	${\mathrm{m}}^3$
$V_4$	Volume at corner 4, where the cold isotherm ends.	${\mathrm{m}}^3$
$T_h$	Temperature of the hot isotherm and of the hot reservoir.	$\mathrm{K}$
$T_c$	Temperature of the cold isotherm and of the cold reservoir.	$\mathrm{K}$
$Q_h$	Heat the gas takes in from the hot reservoir along the hot isotherm.	$\mathrm{J}$
$Q_c$	Heat the gas gives to the cold reservoir along the cold isotherm.	$\mathrm{J}$
$W$	Net work done by the gas in one cycle: the area inside the loop.	$\mathrm{J}$

Figure 2 shows the animation's Carnot loop to scale. Follow it clockwise from corner 1. From 1 to 2 the gas expands along the hot isotherm, at $T_h$, and takes in the heat $Q_h$ from the hot reservoir. From 2 to 3 it keeps expanding along an adiabat; with no heat coming in it does this work out of its own internal energy, so it cools until it reaches $T_c$. From 3 to 4 it is compressed along the cold isotherm, at $T_c$, and gives up the heat $Q_c$ to the cold reservoir. From 4 to 1 a second adiabat compresses it with no heat flow, and the work done on it warms it back to $T_h$.

The corners are at $V_1=1\mathrm{L}$ and 1000 kPa, $V_2=3\mathrm{L}$ and 333 kPa, $V_3=6.46\mathrm{L}$ and 93 kPa, and $V_4=2.15\mathrm{L}$ and 279 kPa; the next section shows where $V_3$ and $V_4$ come from. The two isotherms are the hyperbolas $pV=1000\mathrm{J}$ and $pV=nR{T}_c=600\mathrm{J}$.

Now compare the curves at a corner. An adiabat obeys $p{V}^{\gamma }=\text{const}$ (derived in the next section), where $\gamma$, the ratio of the heat capacities at constant pressure and at constant volume, is 5/3 for a monatomic gas. Differentiating, its slope is $dp/dV=-\gamma p/V$, against $-p/V$ for an isotherm through the same point. So where the two cross, the adiabat falls 5/3 times as steeply. That is why each adiabat in Figure 2 cuts down from the hot isotherm to the cold one, closing the loop.

Read the work off the areas. Under the two expansion strokes, 1 to 2 and 2 to 3, the area is $1098.6+600=1698.6\mathrm{J}$; under the two compressions, 3 to 4 and 4 to 1, it is $659.2+600=1259.2\mathrm{J}$. The shaded loop between them is the net work, $W=1698.6-1259.2=439.4\mathrm{J}$.

The two adiabatic areas are equal, 600 J each, because on an adiabat $Q=0$, so the work equals the change of $U$, and both adiabats change $U$ by $\frac{3}{2}nR(T_h-T_c)=1.5\cdot 2.00\cdot 200=600\mathrm{J}$. The loop runs clockwise, so the gas does more work while expanding at high pressure than is done on it while being compressed at low pressure. Run anticlockwise, the same loop is a refrigerator.

Deriving the Carnot efficiency for an ideal gas

The efficiency of an engine is the fraction of the heat it takes in that comes out as work, $\eta =W/Q_h$. Over one cycle $\Delta U=0$, so $W=Q_h-Q_c$ and $\eta =1-Q_c/Q_h$. Everything therefore hangs on the ratio $Q_c/Q_h$, and the strokes of Figure 2 give it in three steps.

First, the isotherms. For an ideal gas $U$ depends on $T$ only, so along an isotherm $dU=0$ and $\delta Q=\delta W=pdV$. With $p=nRT/V$ and $T$ fixed,
$Q={\int }_{V_i}^{V_f}\frac{nRT}{V}dV=nRT\ln \frac{V_f}{V_i}$
The hot isotherm, from $V_1$ to $V_2$ in Figure 2, takes in $Q_h=nR{T}_h\ln (V_2/V_1)=1000\ln 3=1098.6\mathrm{J}$. The cold one runs as a compression from $V_3$ to $V_4$, so the gas gives out heat; counted as a positive amount, $Q_c=nR{T}_c\ln (V_3/V_4)$.

Second, the adiabats. Along an adiabat $\delta Q=0$, so $dU=-pdV$. Substitute $dU=\frac{3}{2}nRdT$ and $p=nRT/V$, then divide by $nRT$:
$\frac{3}{2}\frac{dT}{T}=-\frac{dV}{V}$
Integrating gives $\frac{3}{2}\ln T+\ln V=\text{const}$, that is $T{V}^{2/3}=\text{const}$. For any ideal gas the same steps give $T{V}^{\gamma -1}=\text{const}$, and replacing $T$ by $pV/nR$ turns this into $p{V}^{\gamma }=\text{const}$, with $\gamma =5/3$ here.

Third, the volumes. The adiabat from 2 to 3 starts at $T_h$ and volume $V_2$ and ends at $T_c$ and volume $V_3$, so $T_h{V}_2^{2/3}=T_c{V}_3^{2/3}$, which gives $V_3/V_2=(T_h/T_c{)}^{3/2}$. The adiabat from 4 to 1 joins the same two temperatures, so $V_4/V_1$ is the same factor. Dividing one ratio by the other,
$\frac{V_3}{V_4}=\frac{V_2}{V_1}$
In the animation the factor is $(5/3{)}^{3/2}=2.15$: it takes 3 L to 6.46 L and 1 L to 2.15 L, the volume steps of the two adiabats in Figure 2.

The two logarithms are therefore equal, $\ln (V_3/V_4)=\ln (V_2/V_1)=\ln 3$, so $Q_c=600\ln 3=659.2\mathrm{J}$, and dividing $Q_h$ by $Q_c$ leaves only the temperatures:
$\frac{Q_h}{T_h}=\frac{Q_c}{T_c}\text{(Carnot)}$
${\eta }_C=1-\frac{T_c}{T_h}$
Both ratios equal $nR\ln 3=2.197J/K$ in the animation, and the efficiency is $1-300/500=40\%$, whatever the volumes or the amount of gas. The exponent $\gamma$ has also dropped out: a diatomic gas would need other volumes but would reach the same efficiency. That is a first hint that the result belongs to no particular substance, and the next two sections prove it.

Reversible engines and the second law

A process is reversible if an arbitrarily small change of conditions can run it backwards, through the same states, with every flow of heat and work reversed. That needs two things: no friction, and heat that crosses only vanishing temperature differences. Every Carnot stroke qualifies. On the isotherms the gas is at the reservoir's own temperature, and on the adiabats no heat flows at all. So the engine can run in reverse, using 439.4 J of work to take 659.2 J from the cold reservoir and deliver 1098.6 J to the hot one.

The second law needs one more ingredient, a statement of what nature forbids. In the form Kelvin gave it, later sharpened by Planck: no cycle can take heat from a single reservoir and turn it into work with no other change. It sounds modest, but the next section shows that it alone fixes the best efficiency any engine can have.

Why no engine beats a reversible one

The argument is a thought experiment. Suppose someone builds an engine $A$ that does better than a Carnot engine between the same two reservoirs. Couple it to a Carnot engine $C$ running backwards and add up what happens.

Legend

$T_h$	Temperature of the hot reservoir.	$\mathrm{K}$
$T_c$	Temperature of the cold reservoir.	$\mathrm{K}$
$A$	Engine claimed to be more efficient than a reversible one.
$C$	Carnot engine, run backwards as a refrigerator.
$Q_h$	Heat that $A$ takes from the hot reservoir and that $C$ gives back to it: the same amount.	$\mathrm{J}$
$Q_A$	Heat that $A$ rejects to the cold reservoir.	$\mathrm{J}$
$Q_c$	Heat that $C$ takes from the cold reservoir, more than $Q_A$.	$\mathrm{J}$
$W_A$	Work that $A$ delivers per cycle.	$\mathrm{J}$
$W_C$	The part of $W_A$ passed on to drive $C$.	$\mathrm{J}$
$W_A-W_C$	Work left over: the heat the cold reservoir loses, $Q_c-Q_A$, turned into work.	$\mathrm{J}$

In Figure 3 the hot reservoir at $T_h=500\mathrm{K}$ is at the top and the cold one at $T_c=300\mathrm{K}$ at the bottom. Vertical arrows carry heat, horizontal arrows carry work. Engine $A$ takes the same $Q_h=1098.6\mathrm{J}$ from the hot reservoir as the Carnot engine would, but is supposed to deliver more work, $W_A=500\mathrm{J}$, an efficiency of 45.5 %. By the first law it then rejects only $Q_A=1098.6-500=598.6\mathrm{J}$ to the cold reservoir. Now pass $W_C=439.4\mathrm{J}$ of that work to $C$, run as a refrigerator: it takes $Q_c=659.2\mathrm{J}$ from the cold reservoir and delivers $Q_h=1098.6\mathrm{J}$ to the hot one.

Follow the arrows and keep the accounts. The hot reservoir gives $Q_h$ to $A$ and gets the same $Q_h$ back from $C$, so it is left exactly as it was. The cold reservoir receives $Q_A$ and gives up $Q_c$, a net loss of $Q_c-Q_A=659.2-598.6=60.6\mathrm{J}$. Meanwhile $W_A-W_C=500-439.4=60.6\mathrm{J}$ of work is left over. Taken together, the two machines turn heat from the cold reservoir alone into work, with no other change. The second law forbids exactly this, so engine $A$ cannot exist: no engine beats a reversible one.

Now let $A$ be reversible too. If it were less efficient than $C$, the same argument with the roles swapped, $C$ driving $A$ backwards, would again give work from one reservoir. So all reversible engines between the same two temperatures share one efficiency, whatever their working substance, and the ideal-gas calculation has already found its value: ${\eta }_C=1-T_c/T_h$.

That universality defines an absolute temperature with no thermometer substance at all: set $T_h/T_c=Q_h/Q_c$ for any reversible engine running between two bodies. The ideal-gas calculation shows that this scale is proportional to the ideal-gas scale, so the two agree once the same unit is chosen. The zeroth law (two bodies each in equilibrium with a third are in equilibrium with each other) is what makes "the temperature of a reservoir" a meaningful label in the first place.

Finally, read $Q_h/T_h=Q_c/T_c$ as a balance. The engine takes the quantity $Q/T$ out of the hot reservoir and passes the same amount to the cold one, unchanged. That quantity is the change of entropy, the subject of the lesson on the second law and entropy. The line $\Delta S_{\mathrm{tot}}=Q_c/T_c-Q_h/T_h=0$ in the animation's Advanced layer records 2.197 J/K leaving the hot side and reaching the cold side, with nothing created.

Worked problem: a cycle without adiabats

What does it cost to break one of Carnot's rules? Keep the animation's gas, reservoirs and isotherms between $V_1=1\mathrm{L}$ and $V_2=3\mathrm{L}$, but replace the adiabats by constant-volume steps: lock the piston and move the cylinder to the other reservoir. Find the work, the efficiency and where the shortfall goes.

Legend

$p$	Pressure of the gas.	$\mathrm{Pa}$
$V$	Volume of the gas.	${\mathrm{m}}^3$
$V_1$	The smaller volume, 1 L, where the left-hand step is made.	${\mathrm{m}}^3$
$V_2$	The larger volume, 3 L, where the right-hand step is made.	${\mathrm{m}}^3$
$T_h$	Temperature of the hot isotherm and of the hot reservoir.	$\mathrm{K}$
$T_c$	Temperature of the cold isotherm and of the cold reservoir.	$\mathrm{K}$
$W$	Net work done by the gas in one cycle: the area inside the loop.	$\mathrm{J}$
$Q_{41}$	Heat taken from the hot reservoir while the gas warms at constant volume, from corner 4 to corner 1.	$\mathrm{J}$
$Q_{23}$	Heat given to the cold reservoir while the gas cools at constant volume, from corner 2 to corner 3.	$\mathrm{J}$

Figure 4 shows the new loop to scale. From 1 to 2 the gas expands along the hot isotherm ($T_h$) and from 3 to 4 it is compressed along the cold one ($T_c$), exactly as before, so the isotherms still give 1098.6 J in at 500 K and 659.2 J out at 300 K. The two vertical steps are new. At constant volume the gas does no work, so the heat it exchanges equals its change of internal energy.

From 2 to 3, at 3 L, the gas cools from 500 K to 300 K and its pressure falls from 333 kPa to 200 kPa; it hands the cold reservoir $Q_{23}=\frac{3}{2}nR\cdot 200=600\mathrm{J}$. From 4 to 1, at 1 L, it must take $Q_{41}=600\mathrm{J}$ from the hot reservoir to warm up again, its pressure rising from 600 kPa to 1000 kPa. The vertical steps add no area, so the net work is the same as Carnot's, but the hot reservoir now supplies $1098.6+600=1698.6\mathrm{J}$:
$W=1098.6-659.2=439.4\mathrm{J},\eta =\frac{439.4}{1098.6+600}=25.9\%$

The loss comes from heat crossing a finite temperature gap, and entropy measures it. On the isotherms the entropy flows cancel exactly as in the Carnot cycle. On the vertical steps they do not. While warming at 1 L the gas gains entropy $\int \frac{3}{2}nRdT/T=\frac{3}{2}nR\ln (5/3)=1.53J/K$, but the hot reservoir, giving its 600 J at a steady 500 K, loses only $600/500=1.20J/K$. While cooling at 3 L the gas loses 1.53 J/K, but the cold reservoir gains $600/300=2.00J/K$. The gas returns to its starting state, so its own changes cancel, and each cycle creates $2.00-1.20=0.80J/K$ of entropy in the reservoirs.

The work lost is $T_c$ times the entropy created, $300\cdot 0.80=240\mathrm{J}$. Check it: a Carnot engine given the same 1698.6 J would deliver $0.40\cdot 1698.6=679.4\mathrm{J}$, and $679.4-439.4=240\mathrm{J}$ exactly. The cure is a regenerator, a porous heat store that takes up the 600 J layer by layer as the gas cools and returns it the same way as the gas warms. Each layer meets gas at nearly its own temperature, so no entropy is created, the hot reservoir supplies only the 1098.6 J of the isotherm, and the efficiency is back to 40 %. This is the idea behind the Stirling engine.

Refrigerators and heat pumps

Run backwards, the Carnot engine pumps heat uphill, from cold to hot. A refrigerator is judged by its coefficient of performance $\mathrm{COP}$, the heat removed from the cold side per joule of work, $Q_c/W$. With $W=Q_h-Q_c$ and the Carnot ratio $Q_h/Q_c=T_h/T_c$, the best possible value is
$\mathrm{COP}=\frac{Q_c}{W}\le \frac{T_c}{T_h-T_c}$
In the animation $659.2/439.4=1.5=300/200$. Used as a heat pump, the same machine delivers $Q_h$ to the warm side: $1098.6/439.4=2.5$ joules of heat per joule of work, the limit $T_h/(T_h-T_c)=500/200$. Values above 1 break no law: the work does not have to create the heat it moves, only lift it.

Why real engines fall short

Real engines lose for reasons beyond friction. Most take in heat over a range of temperatures instead of all at the hottest one. The ideal petrol-engine (Otto) cycle is two adiabats joined by heating and cooling at constant volume. Its efficiency is $\eta =1-r^{1-\gamma }$, where the compression ratio $r$ is the largest volume divided by the smallest: 60 % for $r=10$ with air ($\gamma =1.4$). That is well below the Carnot value of nearly 90 % between its peak gas temperature, around 2500 K, and the 300 K intake air. Real petrol engines, with friction and heat leaks as well, peak near 35–40 %.

Speed matters too. A reversible engine must exchange heat across vanishing temperature differences, so it runs infinitely slowly and delivers no power. An engine limited by how fast heat flows in and out gives its greatest power near $\eta =1-\sqrt{T_c/T_h}$, which is 39 % between 800 K and 300 K, close to what real steam power stations achieve.

Limits of validity

The stroke formulas assume quasi-static changes. Pressure evens out at about the speed of sound, crossing a 10 cm cylinder in about 0.2 ms, well within the 10 ms of one stroke in a car engine at 3000 rpm. Heat diffuses far more slowly, so fast strokes are nearly adiabatic and truly isothermal ones are impractical; real cycles add heat at roughly constant volume or constant pressure instead.

The values $U=\frac{3}{2}nRT$ and $\gamma =5/3$ hold for a dilute monatomic gas. Air near room temperature has $\gamma =1.4$, and in a dense gas $U$ depends on the volume too, yet ${\eta }_C$ is unaffected, as the argument of Figure 3 guarantees. The first law has no known exception, but in relativity energy includes rest energy. Burning 1 kg of petrol releases about 46 MJ, so the products are lighter than the petrol and oxygen by $46\times 10^6/(3.00\times 10^8{)}^2\approx 5\times 10^{-10}\mathrm{kg}$, far too little to weigh.

The laws also describe averages. For a handful of molecules $Q$ and $W$ fluctuate from cycle to cycle, and a single cycle can even beat Carnot. The relative fluctuations scale as $1/\sqrt{N}$ for $N$ molecules, about $3\times 10^{-12}$ for the $1.4\times 10^{23}$ atoms of the animation's gas, so for anything visible the laws hold exactly. The third law makes absolute zero unreachable, so $T_c$ is never zero and ${\eta }_C$ always stays below 1.

Where this leads

The Second Law of Thermodynamics and Entropy turns $Q/T$ into entropy, proves the Clausius inequality and explains, by counting molecular arrangements, why heat flows only downhill. Kinetic Theory of Gases gives the molecular reasons for $U=\frac{3}{2}nRT$ and for adiabatic heating. Steam engines rely on Phase Changes and Latent Heat: boiling and condensing happen at a fixed temperature, so their heat exchange is naturally isothermal. Conduction, Convection and Radiation sets how fast heat reaches an engine, and so its power; Electric Power and Energy at Home follows the energy from the power station to the socket; and Work, Energy and Power is where the conservation of energy began.