Temperature and Heat

Beginner

Step 1. How hot is something? A thermometer answers with a number: its temperature. On the Celsius scale, water freezes at 0 °C and boils at 100 °C. Scientists also use the kelvin scale, which starts at absolute zero (about −273 °C), the coldest temperature possible; one kelvin is the same step as one degree Celsius.

Step 2. What does the thermometer really measure? Everything is made of tiny particles that never stop jiggling. Temperature tells you how hard, on average, each particle jiggles: the hotter the object, the faster its particles move.

Step 3. Temperature is not the same as the amount of energy. A bathtub of warm water and a teaspoon of the same water have the same temperature, but the bathtub holds thousands of times more jiggling energy, because it contains thousands of times more particles.

Step 4. Heat is energy on the move. When a hot object touches a cold one, its fast particles bump into the slower ones and share some of their motion. Energy flows by itself from hot to cold, never the other way, until both reach the same temperature: thermal equilibrium. In the animation, the hot block's particles jiggle harder at first; the heat arrow shrinks as the thermometers draw level and disappears when both blocks are equally warm.

Step 5. Some materials are much harder to warm than others. A kilogram of water needs about 4200 joules to warm by one degree; a kilogram of iron needs only about 450. That is why sand can burn your feet while the sea stays cool.

Check yourself: a sparkler's sparks are hotter than 1000 °C, yet they barely hurt when they land on your hand. Why? Answer: each spark is tiny, so although it is very hot, it carries very little energy.

Intermediate

Temperature, heat and their units

Temperature is what a thermometer reads. On the Celsius scale water freezes at 0 °C and boils at 100 °C. The kelvin scale (unit K) has steps of the same size but starts at absolute zero, −273.15 °C, the lowest temperature possible:
$T(\mathrm{K})=T({}^{\circ }\mathrm{C})+273.15$

A temperature change $\Delta T$ therefore has the same value on both scales: from 20 °C to 30 °C is a change of 10 K. Only the kelvin scale starts from a true zero, so use kelvins whenever temperatures are multiplied, divided or raised to a power.

Heat $Q$ is energy, in joules (J), that passes from one body to another because their temperatures differ. A body does not contain heat: it contains internal energy $U$, the energy of the random motion of its particles and of the forces between them. Heat, like work, is energy in transit, positive when it flows into a body and negative when it flows out.

Specific heat capacity

As long as nothing melts, boils or reacts, the heat needed to change a body's temperature, and the heat capacity of the body, are

• $Q=mc\Delta T$
• $C=mc$

Here $m$ is the mass, $\Delta T$ the final temperature minus the initial one, and $c$ the specific heat capacity: the energy that warms 1 kg of the material by 1 K, in J/(kg·K). The heat capacity $C=mc$ of the whole body is in J/K. When a body cools, $\Delta T$ and $Q$ are both negative.

Typical values, in J/(kg·K), are 4180 for water, 2100 for ice, 900 for aluminium, about 800 for dry sand, 450 for iron, 385 for copper and 130 for lead. Kilogram for kilogram, water needs about nine times as much energy as iron for the same warming. That is why the sea warms and cools slowly.

Heat balance and the final temperature

When a hot body and a cold body touch inside good insulation, heat flows until both reach a common temperature $T$, and the heat lost by the hot body equals the heat gained by the cold one:
$m_1{c}_1(T_1-T)=m_2{c}_2(T-T_2)$
Solving for $T$, the equilibrium temperature, gives
$T_{\mathrm{eq}}=\frac{m_1{c}_1{T}_1+m_2{c}_2{T}_2}{m_1{c}_1+m_2{c}_2}$

This is an average of the starting temperatures, weighted by heat capacity: it lies between them, closer to the body with the larger $mc$. A weighted average does not depend on where the zero of the scale lies, so the formula works in °C as well as in kelvins. A body with twice the heat capacity changes its temperature half as much.

Worked example: the blocks in the animation

A 1 kg block of iron ($c_1=450J/(\mathrm{kg}\cdot \mathrm{K})$, so $C_1=450J/K$) at 327 °C touches a 1 kg block of aluminium ($c_2=900J/(\mathrm{kg}\cdot \mathrm{K})$, $C_2=900J/K$) at 27 °C. With no losses, heat lost equals heat gained:
$450(327-T)=900(T-27)$
$1350T=147150+24300=171450,T=127{}^{\circ }\mathrm{C}$

The iron cools by 200 K, while the aluminium warms by only 100 K because its heat capacity is twice as large. The heat that crossed the contact is $Q=450\cdot 200=900\cdot 100=90000\mathrm{J}=90\mathrm{kJ}$. These are the final values on the readouts: $T_1$ and $T_2$ meet at 127 °C while $Q$ climbs to 90 kJ.

The readouts also show the exchange slowing down: after 1 s the iron reads about 217 °C, the aluminium reads about 82 °C, and the heat that has crossed is about 50 kJ, so more than half of the heat crosses in the first second. The orange-red arrow labelled $dQ/dt$, the rate of heat flow in watts, has a length proportional to $T_1-T_2$. It is half as long when the readouts reach about 227 °C and 77 °C, and it fades as the thermometers draw level, the iron's column always falling twice as fast as the aluminium's rises.

Worked example: running a bath

You want an 80 kg bath at 38 °C, mixing hot water at 60 °C with cold water at 12 °C. How much hot water do you need, and how much energy did the boiler spend heating it? Ignore losses.

Both are water, so $c$ cancels: $m_h(60-38)=m_c(38-12)$. With $m_h+m_c=80\mathrm{kg}$, this gives $m_h=80\cdot 26/48\approx 43.3\mathrm{kg}$ of hot water and 36.7 kg of cold, more hot than cold because 38 °C is nearer to 60 °C. The boiler warmed the hot water from 12 °C to 60 °C:
$Q=mc\Delta T=43.3\cdot 4180\cdot 48\approx 8.7\times 10^6\mathrm{J}$

That is 8.7 MJ, about 2.4 kWh (1 kWh is 3.6 MJ), or nearly 50 minutes of a 3 kW heater.

How fast heat flows: the shrinking arrow

The heat balance gives the end point, not the time it takes. In the animation heat crosses the contact at $dQ/dt=k(T_1-T_2)$, with a thermal conductance $k=240W/K$: 72 kW at first, then less and less as the gap closes. Because the flow is proportional to the gap, the gap shrinks by the same fraction in equal times, halving about every 0.87 s, and never quite reaches zero.

A body cooling in a large room at temperature $T_{\mathrm{env}}$ follows the same rule, Newton's law of cooling:
$\frac{dT}{dt}=-\frac{T-T_{\mathrm{env}}}{\tau }$
The left side is the rate of change of temperature, and the time constant $\tau$ grows with the body's heat capacity. The excess decays as $T-T_{\mathrm{env}}=(T_0-T_{\mathrm{env}})e^{-t/\tau }$: tea at 80 °C in a 20 °C room whose excess halves every 12 minutes reads 50 °C after 12 minutes, 35 °C after 24 minutes and 27.5 °C after 36 minutes.

Common mistakes

• Heat is not stored in a body, and cold does not flow: the 90 kJ in the animation is energy that passed from the iron to the aluminium.
• Use the temperature change, not the final temperature. Warming a 0.5 kg aluminium pan from 20 °C to 120 °C takes $0.5\cdot 900\cdot 100=45\mathrm{kJ}$, not 54 kJ.
• Equal heat does not mean an equal change of temperature. The same 90 kJ cools the iron by 200 K but warms the aluminium by only 100 K.
• Celsius temperatures cannot be compared as ratios: 54 °C is not twice as hot as 27 °C, since 327 K is only 9 % more than 300 K.
• A metal handle feels colder than a wooden one at the same 20 °C only because metal, which conducts heat hundreds of times better, draws heat from your skin much faster: skin senses heat loss, not temperature.

Where you meet it

Water's large specific heat gives coastal towns milder winters and cooler summers, and the oceans have absorbed about nine tenths of the extra energy trapped by greenhouse gases. Engines are cooled with water, storage heaters keep warmth in dense bricks, and food energy is measured by burning food in a calorimeter; a food Calorie is 4.18 kJ.

Heat that melts or boils instead of warming is the subject of Phase Changes and Latent Heat, and how heat travels that of Conduction, Convection and Radiation. Thermal Expansion shows what warming does to size, Kinetic Theory of Gases explains temperature as molecular motion, and First Law of Thermodynamics and Heat Engines joins heat and work in one energy balance.

Advanced

Temperature as a condition of equilibrium

Before temperature can be measured it has to be defined, and the definition comes from an experiment. Put two bodies in thermal contact, so that they can exchange energy but not matter, and wait. For a while their properties change: lengths, pressures and electrical resistances drift. Then the changes stop. The bodies are now in thermal equilibrium, and we say, by definition, that they have the same temperature.

The zeroth law of thermodynamics adds one fact: two bodies that are each in equilibrium with a third are in equilibrium with each other. This is what makes a thermometer trustworthy. The thermometer is the third body: if it gives the same reading in contact with A and in contact with B, then A and B would not change if they touched. For this to work, its heat capacity must be small enough not to disturb the temperature it measures.

This definition says when two temperatures are equal, but not how to number them. Mercury, resistance and thermocouple thermometers agree at their calibration points but differ slightly in between, because each relies on a property of its own material. A dilute gas removes this dependence: at constant volume its pressure is proportional to a temperature that is the same for every gas at low enough density. That temperature is the kelvin scale. Since 2019 the kelvin has been fixed by giving the Boltzmann constant an exact value, which ties every temperature $T$ to an energy $k_{B}T$:
$k_B=1.380649\times 10^{-23}J/K$

What collisions share out

Equal temperature also has a meaning at the scale of molecules: it is the condition under which collisions stop moving energy, on average, from one body to another. A thought experiment shows why.

Take two different gases in a box, separated by a piston that slides freely. The piston soon settles where the pressures on its two faces balance, but it never stands quite still: it trembles under the individual molecular impacts. Through this trembling, the side whose molecules strike harder on average passes energy to the other side. The trade stops only when the mean kinetic energy per molecule is the same on both sides. In a mixture, collisions do the same job directly, molecule to molecule. So in equilibrium heavy molecules move more slowly than light ones, yet all have the same average $\frac{1}{2}m{v}^2$.

Classical statistical mechanics turns this into the equipartition theorem. In equilibrium, every term of a particle's energy that is quadratic in one velocity component or one coordinate, such as $\frac{1}{2}m{v}_x^2$, has the same average, $\frac{1}{2}{k}_{B}T$. A molecule of a monatomic gas has three such terms, one kinetic term per direction, so its mean kinetic energy is $\frac{3}{2}{k}_{B}T$. An atom in a solid has three kinetic terms and also three potential ones, because its bonds pull it back towards its rest position along each direction; its mean energy is therefore $3k_{B}T$.

For one mole of a simple solid, with $N_A$ atoms and $R=N_A{k}_B$, the energy is $3RT$, so the molar heat capacity is the same for every element:
$C_m=3R\approx 25J/(\mathrm{mol}\cdot \mathrm{K})$
This is the Dulong–Petit law, found by experiment in 1819. Divided by the molar masses of iron and aluminium, 0.0558 and 0.0270 kg/mol, $3R$ gives specific heat capacities of about 447 and 924 J/(kg·K), close to the 450 and 900 used in the animation. Aluminium's value is about twice iron's because its atoms are about half as heavy: a kilogram of aluminium holds about twice as many atoms, and each stores the same energy.

How far an atom jiggles

Equipartition says how much energy an atom's vibration holds. Figure 1 turns that energy into a distance, the size of the atom's jiggle, and shows how it grows with temperature.

Legend

$U$	Elastic energy of one atom in its bonds, $\frac{1}{2}{k}_s{x}^2$, where $k_s$ is the stiffness of the bonds.	$\mathrm{J}$
$x$	Displacement of the atom from its rest position, along one direction.	$\mathrm{m}$
$\frac{1}{2}{k}_B{T}_1$	Mean elastic energy along one direction at the higher temperature $T_1$, drawn as a level.	$\mathrm{J}$
$\frac{1}{2}{k}_B{T}_2$	Mean elastic energy along one direction at the lower temperature $T_2$, drawn as a level.	$\mathrm{J}$
$x_1$	Where the level $\frac{1}{2}{k}_B{T}_1$ meets the well: the root-mean-square displacement, the typical size of the jiggle, at $T_1$.	$\mathrm{m}$
$x_2$	The same at $T_2$; smaller than $x_1$ by a factor $\sqrt{2}$ when $T_1=2T_2$.	$\mathrm{m}$

Figure 1 follows one atom of a solid moving along one direction, with displacement $x$ from its rest position. For small displacements its bonds act like a spring of stiffness $k_s$, so its elastic energy is $U=\frac{1}{2}{k}_s{x}^2$: the parabolic well in the figure. (Here $U$ is the energy of this one atom in its bonds, not the internal energy of the block.) By equipartition the average of $U$ is $\frac{1}{2}{k}_{B}T$. Each horizontal line is drawn at that average for one temperature: the upper line at $\frac{1}{2}{k}_B{T}_1$ for the hotter body, the lower one at $\frac{1}{2}{k}_B{T}_2$ for the colder body, with $T_1=2T_2$ as in the animation.

Now look at where each line meets the well. There $\frac{1}{2}{k}_s{x}^2=\frac{1}{2}{k}_{B}T$, so the line crosses the well at $x=\pm \sqrt{k_{B}T/k_s}$, one point on each side of the centre. The brackets under the axis measure the crossing from the centre: $x_1$ for $T_1$ and $x_2$ for $T_2$. These distances have a physical meaning. Since the mean of $\frac{1}{2}{k}_s{x}^2$ is $\frac{1}{2}{k}_{B}T$, the mean of $x^2$ is $k_{B}T/k_s$, so each crossing is exactly the root-mean-square displacement: the typical size of the jiggle at that temperature.

The shape of the well decides how that size grows. Because $U$ rises as $x^2$, raising the line to twice the height pushes the crossing out by only a factor $\sqrt{2}$: in the figure $x_1=\sqrt{2}{x}_2$. The jiggle grows as $\sqrt{T}$, not as $T$. The animation draws its particles by this rule: the iron's, at 600 K, jiggle $\sqrt{2}\approx 1.4$ times as widely as the aluminium's, at 300 K.

Heat, work and the first law

Temperature says which way energy will flow; heat is the name of that flow. Heat was once pictured as a fluid, caloric, stored in bodies. In the 1840s James Joule showed that falling weights turning a paddle wheel in water warm it just as heating does, at a fixed rate of about 4.2 J of work for every calorie. Heat is therefore not a substance but energy in transfer, and energy conservation that includes it is the first law of thermodynamics:
$\Delta U=Q-W$

Here $U$ is the internal energy of the system, $Q$ the heat it receives and $W$ the work it does on its surroundings, the sign convention of the lesson First Law of Thermodynamics and Heat Engines. If $W$ counts the work done on the system instead, the law reads $\Delta U=Q+W$.

The two kinds of transfer differ in how they happen. Work is energy passed on by a macroscopic force acting through a distance, like a piston pushing on a gas. Heat is energy passed on by microscopic collisions or by radiation, driven by a temperature difference. Only $U$ is a property of the state of a body. A warm block could have been warmed by a flame, which is heat, or by rubbing, which is work, and nothing in the block records which; so "the heat contained in a body" means nothing.

For solids and liquids, the work done against the atmosphere as they expand is negligible, so the heat they receive, $Q=C\Delta T$, is in effect their change of internal energy $\Delta U$. Gases differ: an ideal gas heated at constant pressure also does work as it expands, so its molar heat capacity at constant pressure exceeds the one at constant volume by exactly $R$.

Two blocks in contact: an exponential approach

The first law fixes where the exchange ends; to follow it in time we need a model of how fast heat crosses the contact. Treat each block as uniform, with one temperature throughout: the hot block has heat capacity $C_1$ and temperature $T_1(t)$, the cold one $C_2$ and $T_2(t)$. Let the contact carry a heat flow proportional to the temperature difference, $k(T_1-T_2)$, where $k$ is the thermal conductance of the contact, in W/K. Each block's energy changes only through this flow, so
$C_1\frac{d{T}_1}{dt}=-k(T_1-T_2),C_2\frac{d{T}_2}{dt}=k(T_1-T_2)$

Two results follow. First, adding the equations gives $d(C_1{T}_1+C_2{T}_2)/dt=0$: what one block loses, the other gains, so $C_1{T}_1+C_2{T}_2$ stays at its starting value. At the end both temperatures are equal, and this fixes the final temperature:
$T_{\mathrm{eq}}=\frac{C_1{T}_1(0)+C_2{T}_2(0)}{C_1+C_2}$

Second, dividing each equation by its heat capacity and subtracting the second from the first gives an equation for the gap $\Delta T=T_1-T_2$ alone:
$\frac{d\Delta T}{dt}=-k\left(\frac{1}{C_1}+\frac{1}{C_2}\right)\Delta T$
The gap shrinks at a rate proportional to its own size, which is the mark of exponential decay. Starting from $\Delta T_0$,
$\Delta T=\Delta T_0{e}^{-t/\tau },\tau =\frac{C_1{C}_2}{k(C_1+C_2)}$
The time constant $\tau$ is long when the bodies are large or the contact is poor. The combination $C_1{C}_2/(C_1+C_2)$ is always smaller than either heat capacity, so the smaller body sets the pace.

The animation uses 1 kg of iron at 600 K (327 °C), so $C_1=450J/K$, and 1 kg of aluminium at 300 K (27 °C), so $C_2=900J/K$. Then $T_{\mathrm{eq}}=(450\cdot 600+900\cdot 300)/1350=400\mathrm{K}$, or 127 °C, and the iron passes $450\cdot 200=90\mathrm{kJ}$ to the aluminium. With $C_1{C}_2/(C_1+C_2)=300J/K$ and $k=240W/K$, the time constant is $\tau =300/240=1.25\mathrm{s}$ ($1/\tau =0.8{\mathrm{s}}^{-1}$). The flow starts at $k\Delta T_0=240\cdot 300=72\mathrm{kW}$, the gap halves every $\tau \ln 2\approx 0.87\mathrm{s}$, and it is down to 1 K after $\tau \ln 300\approx 7.1\mathrm{s}$.

Reading the two curves

The solution is easiest to grasp as a picture. Figure 2 shows where the two temperatures end, how fast they start, and what $\tau$ means on the graph.

Legend

$T$	Temperature of a block.	$\mathrm{K}$
$t$	Time since the blocks were put in contact.	$\mathrm{s}$
$T_1$	Temperature of the hot block, the iron; it falls from its starting value.	$\mathrm{K}$
$T_2$	Temperature of the cold block, the aluminium; it rises from its starting value.	$\mathrm{K}$
$T_{eq}$	Common final temperature, the level both curves approach.	$\mathrm{K}$
$\Delta T_0$	Starting temperature difference $T_1-T_2$; the gap then shrinks as $e^{-t/\tau }$.	$\mathrm{K}$
$\tau$	Time constant: the time at which both starting tangents reach $T_{\mathrm{eq}}$.	$\mathrm{s}$

Figure 2 draws the two curves of the animation's expert graph with its numbers. The orange-red curve $T_1$ starts at the iron's 327 °C and falls; the blue curve $T_2$ starts at the aluminium's 27 °C and rises. The bracket on the left spans the starting gap $\Delta T_0$, 300 K.

First, see where the curves go. Both bend onto the green dashed line $T_{\mathrm{eq}}$, at 127 °C. This line cuts the bracket in the ratio 2 : 1, and it keeps cutting the shrinking gap in the same ratio: because $C_1{T}_1+C_2{T}_2$ always equals $(C_1+C_2)T_{\mathrm{eq}}$, we have $C_1(T_1-T_{\mathrm{eq}})=C_2(T_{\mathrm{eq}}-T_2)$, so the iron, with half the heat capacity, is always twice as far from $T_{\mathrm{eq}}$. At every moment $T_1-T_{\mathrm{eq}}=\frac{2}{3}\Delta T$ and $T_{\mathrm{eq}}-T_2=\frac{1}{3}\Delta T$: the iron's curve is the aluminium's mirrored in the dashed line and stretched to twice the height.

Next, see how they start. The two grey dashed lines are the tangents at $t=0$, and their slopes are the starting rates: $-k\Delta T_0/C_1=-160K/s$ for the iron and $+k\Delta T_0/C_2=+80K/s$ for the aluminium. Both tangents reach the line $T_{\mathrm{eq}}$ at the same instant, $t=\tau$, marked on the time axis: the iron would fall its 200 K, and the aluminium rise its 100 K, in 1.25 s.

This is what $\tau$ means: the time a temperature would need to reach $T_{\mathrm{eq}}$ if it kept its starting rate. In fact the rate falls in proportion to the remaining gap, so after $\tau$ about 37 % of the gap is left, and after $4\tau$ only about 2 %.

If the second body is huge ($C_2\to \infty$, a room or a lake), its temperature hardly moves from $T_{\mathrm{env}}$, $\tau$ becomes $C_1/k$, and the first equation becomes Newton's law of cooling:
$\frac{dT}{dt}=-\frac{T-T_{\mathrm{env}}}{\tau }$

Why heat runs from hot to cold

Nothing in the first law sets the direction of the flow: energy would be conserved just as well if the 90 kJ went back from the aluminium into the iron. The direction comes from the second law of thermodynamics: the total entropy of an isolated system cannot decrease. A body at a uniform temperature $T$ that receives a small heat $\delta Q$ gains entropy $\delta Q/T$. Figure 3 applies this rule to one small transfer between the blocks.

Legend

$T_1$	Temperature of the hot block.	$\mathrm{K}$
$T_2$	Temperature of the cold block.	$\mathrm{K}$
$\delta Q$	A small amount of heat passing from the hot block to the cold one.	$\mathrm{J}$
$\delta Q/T_1$	Entropy lost by the hot block.	$J/K$
$\delta Q/T_2$	Entropy gained by the cold block.	$J/K$
$dS$	Entropy created by the transfer: the overhang of the lower bar, $\delta Q(1/T_2-1/T_1)$.	$J/K$

Figure 3 follows one small amount of heat $\delta Q$ from the hot block, at $T_1$, to the cold one, at $T_2$; it is small enough that neither temperature changes during the transfer. The hot block loses the entropy $\delta Q/T_1$, the upper bar; the cold block gains $\delta Q/T_2$, the lower bar. The heat is the same on both sides, but on the cold side it is divided by a smaller temperature, so that bar is longer: twice as long here, where $T_1=2T_2$. The part of the lower bar that overhangs the upper one, $dS$, is the entropy that the transfer creates:
$dS=\delta Q\left(\frac{1}{T_2}-\frac{1}{T_1}\right)$

For $\delta Q>0$ this is positive exactly when $T_1>T_2$. Heat flowing from hot to cold therefore raises the total entropy, and the reverse flow would lower it, which the second law forbids. At the molecular level the reverse flow breaks no law of motion; for bodies of everyday size it is just overwhelmingly improbable.

For the whole exchange in the animation, add up such steps. Each block's entropy changes by $dS=CdT/T$, with $T$ in kelvins, and integrating from the starting temperatures to $T_{\mathrm{eq}}=400\mathrm{K}$ gives
$\Delta S=C_1\ln \frac{T_{\mathrm{eq}}}{T_1(0)}+C_2\ln \frac{T_{\mathrm{eq}}}{T_2(0)}=450\ln \frac{400}{600}+900\ln \frac{400}{300}\approx 76J/K$
The iron loses 182 J/K, but the aluminium gains 259 J/K: it receives the same 90 kJ, but at lower temperatures, so each joule brings it more entropy.

The second law also gives temperature its deepest definition. Take a reversible engine that takes heat $Q_h$ from a reservoir at $T_h$ and rejects heat $Q_c$ to a reservoir at $T_c$. Whatever its working substance, $Q_h/Q_c=T_h/T_c$, so this ratio defines a temperature scale that depends on no material, and it coincides with the ideal-gas kelvin scale. In statistical physics the same idea reads $1/T=\partial S/\partial U$ at fixed volume: the inverse temperature is the entropy a body gains per joule it receives. That is why heat flows towards the colder body: there each joule brings more entropy.

Worked problem: an engine between the blocks

Direct contact delivers no work, although the temperature difference between the blocks could yield some. Suppose that, instead of touching, the two blocks drive a reversible engine that takes heat from the iron, rejects heat to the aluminium and delivers work, until both reach a common temperature $T_f$. The iron starts at $T_1=600\mathrm{K}$ and the aluminium at $T_2=300\mathrm{K}$. What is $T_f$, and how much work is delivered?

Legend

$T$	Temperature of a block.	$\mathrm{K}$
$s$	Entropy that has left the iron so far; the same amount has entered the aluminium.	$J/K$
$T_1$	Starting temperature of the iron, 600 K; the iron cools along the orange-red curve.	$\mathrm{K}$
$T_2$	Starting temperature of the aluminium, 300 K; it warms along the blue curve.	$\mathrm{K}$
$T_f$	Common final temperature, where the two curves meet.	$\mathrm{K}$
$W$	Work delivered by the engine: the area between the curves.	$\mathrm{J}$

Figure 4 follows the engine on a temperature–entropy diagram. Let $s$ be the entropy that has left the iron so far; it runs along the horizontal axis. A reversible engine creates no entropy, so the same $s$ has entered the aluminium. Since $dS=CdT/T$ for each block, the iron, losing $s$, cools along $T_1{e}^{-s/C_1}$, the orange-red curve, while the aluminium, gaining $s$, warms along $T_2{e}^{s/C_2}$, the blue one. The curves meet at the dashed level $T_f$; there no temperature difference is left and the engine stops.

Now read off the work. In a small step $ds$ the iron gives up a heat equal to its own temperature times $ds$, and the aluminium takes in its own, lower temperature times $ds$. The engine turns the difference into work, which on the diagram is a thin vertical strip between the curves. Adding the strips, the work $W$ is the shaded area between the curves, which narrows to a point as the temperatures approach each other.

The curves meet when the entropy changes of the two blocks cancel, since the total entropy does not change: $C_1\ln (T_f/T_1)+C_2\ln (T_f/T_2)=0$. Solving for $\ln T_f$ gives
$T_f=T_1^{C_1/(C_1+C_2)}{T}_2^{C_2/(C_1+C_2)}=600^{1/3}\cdot 300^{2/3}=300\cdot 2^{1/3}\approx 378\mathrm{K}$
or about 105 °C. By then $s=C_1\ln (T_1/T_f)\approx 208J/K$ has passed from one block to the other. No heat leaves the pair except through the engine, so the work is the internal energy the blocks lose together:
$W=C_1(T_1-T_f)-C_2(T_f-T_2)\approx 99.9-70.2=29.7\mathrm{kJ}$

The engine takes 99.9 kJ from the iron, more than the 90 kJ of direct contact, and turns about 30 % of it into work; its momentary efficiency, $1-T_2/T_1$ for the current temperatures, falls from 50 % at the start to zero at the end. The blocks finish 22 K below the 400 K of direct contact, because the reversible end point is the weighted geometric mean of the starting temperatures, while contact gives the weighted arithmetic mean, which is always higher. Any temperature difference can yield work; letting heat leak straight across wastes it and creates the 76 J/K found above.

Real blocks and the limits of the model

The two-block model assumes that each block stays uniform, which needs heat to spread inside a block faster than it crosses the contact. Inside a body, heat flows down the temperature gradient, following Fourier's law of conduction:
$\vec{q}=-\kappa \nabla T$
Here $\vec{q}$ is the heat flux in W/m² and $\kappa$ the thermal conductivity, about 80 W/(m·K) for iron. Combined with energy conservation, it gives the diffusion equation $\partial T/\partial t=\alpha {\nabla }^{2}T$, with the thermal diffusivity $\alpha =\kappa /(\rho c)\approx 2.3\times 10^{-5}{\mathrm{m}}^2/s$ for iron, where $\rho$ is the density.

That diffusivity tests the assumption of uniform blocks. Heat needs a time of order $L^2/\alpha$ to spread across a distance $L$: about 110 s across a 1 kg iron cube of side $L\approx 5\mathrm{cm}$. That is far longer than the animation's $\tau$ of 1.25 s, so real blocks would not stay uniform and would take minutes to settle; the animation idealises them.

Hot bodies also radiate. A surface of area $A$ and emissivity $\epsilon$ emits the power
$P=\epsilon \sigma A{T}^4$
with $\sigma =5.67\times 10^{-8}W/({\mathrm{m}}^2\cdot {\mathrm{K}}^4)$, and absorbs in return from its surroundings at $T_0$, so its net loss is $\epsilon \sigma A(T^4-T_0^4)$. The iron's free faces, 0.013 m² with $\epsilon \approx 0.6$, lose a net 50 W or so at 600 K to surroundings at 300 K, so the insulation is only approximate. For small differences $T^4-T_0^4\approx 4T_0^3\Delta T$, and radiation behaves like Newton's law of cooling; but between 600 K and 300 K the flux is nearly four times this linear estimate.

Heat capacities are not constant either: iron's $c$ rises by more than a quarter between 300 K and 600 K. At low temperatures equipartition fails. Vibration energies are quantised, and once $k_{B}T$ is small compared with the spacing of their levels the vibrations freeze out; their heat capacity then falls as $T^3$ near absolute zero. Stiff diamond shows this even at room temperature, with about 510 J/(kg·K) instead of the Dulong–Petit 2080. Finally, temperature itself needs many particles near equilibrium: a single molecule has an energy, not a temperature.

Where this leads

Kinetic Theory of Gases turns $\frac{3}{2}{k}_{B}T$ into pressure and molecular speeds. First Law of Thermodynamics and Heat Engines builds engines on the energy balance, and The Second Law of Thermodynamics and Entropy generalises the 76 J/K found here. Conduction, Convection and Radiation develops Fourier's law and radiation, Phase Changes and Latent Heat treats heat that changes the state rather than the temperature, and Thermal Expansion shows what warming does to size. The failure of equipartition opened the quantum era, as Blackbody Radiation and the Quantum explains.