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The Clausius relation $dS = dQ/T$: entropy changes of reservoirs, phase changes and heating, entropy created when heat crosses a temperature difference, and the third law.
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By the end of this lesson you will be able to find entropy changes from heat for a reservoir, a phase change and a temperature change, add them to find the entropy created in an irreversible process, and explain how absolute entropies are measured from the third law.
You know temperature as $1/T = \partial S/\partial U$, entropy as $k\ln\Omega$, and how to find heat for a temperature change, $Q = mc\Delta T$, and for a phase change, $Q = mL$. This lesson connects heat to entropy directly, so that entropy changes can be found from measurements rather than by counting microstates.
| Term | What it means |
|---|---|
| Clausius relation | $dS = dQ/T$ for heat added reversibly at temperature $T$. |
| Reservoir | A body so large that heat added to or taken from it does not change its temperature. |
| Reversible process | A process slow and gentle enough that it creates no entropy; run backward, it restores everything. |
| Entropy created | $S_{\text{created}} = \Delta S_{\text{total}}$ in an irreversible process; always positive. |
| Isentropic | At constant entropy; a reversible adiabatic process is isentropic. |
| Third law | The entropy of a system approaches a constant, usually zero, as $T \to 0$. |
| Absolute entropy | An entropy measured from absolute zero, $S(T) = \int_0^T C\,dT'/T'$. |
Hold a system's volume fixed and add a little heat $dQ$. No work is done, so $dU = dQ$, and the definition of temperature, $1/T = \partial S/\partial U$, gives
$$dS = \frac{dQ}{T}.$$
This Clausius relation holds for any process slow enough that the system stays in equilibrium — a reversible process — and it lets entropy changes be measured rather than counted. Three cases cover most problems.
$$\Delta S = \int_{T_i}^{T_f}\frac{C\,dT}{T} = C\ln\frac{T_f}{T_i}.$$
Now put heat $Q$ through a rod from a hot reservoir at $T_h$ to a cold one at $T_c$. The hot reservoir loses entropy $Q/T_h$; the cold one gains $Q/T_c$, which is larger because $T_c$ is smaller. So
$$\Delta S_{\text{total}} = \frac{Q}{T_c} - \frac{Q}{T_h} > 0.$$
Energy is conserved — the same $Q$ leaves one and enters the other — but entropy is created. That is the signature of an irreversible process, and it is why heat never flows the other way on its own: the total entropy would have to fall.
Another way: picture
Think of heat as money and entropy as the number of people who can share it. A dollar moving from a rich town to a poor one is the same dollar, but in the poor town it makes a bigger difference. Heat moving from a hot body, where energy is plentiful, to a cold one, where it is scarce, adds more entropy to the cold body than it removes from the hot one. The dollar is conserved; the difference it makes is not.
Another way: steps
Checks. Every temperature must be in kelvin: $\ln(80/20)$ in Celsius gives $1.39$ where $\ln(353/293)$ gives $0.186$. The total change for an isolated set of parts must be positive, or zero only for an idealized reversible process. A part that cools must lose entropy and one that warms must gain it. And an entropy change in joules per kelvin for everyday amounts of material should be tens to thousands, not millions: melting a kilogram of ice gains about $1220$ J/K.
The relation $dS = C\,dT/T$ gives a way to measure the entropy of a substance outright. Measure its heat capacity from as close to absolute zero as possible up to the temperature of interest, divide by $T$, and integrate; at each phase change, add $L/T$. The starting point is fixed by the third law of thermodynamics: as $T \to 0$, the entropy approaches a constant, zero for a perfect crystal with a single lowest-energy state, since then $\Omega = 1$.
For helium gas at $300$ K this chain — the solid's heat capacity, its melting, the liquid, its boiling, the gas warming to room temperature — gives $126.2$ J/(mol K), which is the number the Sackur-Tetrode equation of the entropy lesson predicted by counting. Two routes, one with a calorimeter and a thermometer and one with Planck's constant and a formula for multiplicity, agree to a tenth of a percent. Tables of "standard molar entropies" used throughout chemistry, such as $69.9$ J/(mol K) for liquid water at $298$ K, are measured exactly this way.
The third law also implies that heat capacities must go to zero as $T \to 0$. If $C$ stayed finite, $\int C\,dT/T$ would diverge at the lower limit, giving every substance an infinite entropy — the frozen modes of the Einstein solid are not an accident but a requirement.
The entropy created in an irreversible process measures an opportunity lost. Heat $Q$ leaking from $T_h$ to $T_c$ could instead have driven an engine; the most work it could have produced is $Q(1 - T_c/T_h)$, as the lesson on engines will show. That lost work equals $T_c$ times the entropy created:
$$W_{\text{lost}} = T_c\left(\frac{Q}{T_c} - \frac{Q}{T_h}\right) = T_c\,\Delta S_{\text{created}}.$$
Engineers use this in exergy or second-law analyses of power plants: they add up the entropy each component creates — friction in a turbine, heat crossing a large temperature difference in a boiler, mixing of hot and cold streams — and multiply by the temperature of the surroundings to find where the plant's potential work is being destroyed. It is usually the boiler, where combustion gases at over $1500$ K heat steam at a few hundred kelvin, that wastes the most.
The same final state can be reached reversibly or irreversibly, and the difference shows up only in the surroundings. Heat a kilogram of water from $20$ °C to $80$ °C by plunging it into a single reservoir at $80$ °C: the water gains $780$ J/K as always, but the reservoir loses $Q/T = 251\,000/353 = 711$ J/K, so $69$ J/K is created. Heat it instead through a long series of reservoirs, each a fraction of a degree warmer than the last, and each step transfers heat across an almost zero temperature difference. The reservoirs then lose, in total, almost exactly the $780$ J/K the water gains, and almost nothing is created.
The water's entropy change is identical in both cases, because it depends only on the water's start and end states. What changes is how much entropy the process created, and that is fixed by how large the temperature differences were along the way. Real heat exchangers, from car radiators to the boilers of power plants, are designed to keep those differences small where it matters, because every kelvin of difference across which heat flows is work that can never be recovered.
A coal or nuclear power plant rejects heat to a river, a lake or a cooling tower. A $1000$ MW plant with a thermal efficiency of about $33$ percent burns fuel at $3000$ MW and rejects about $2000$ MW, $2 \times 10^{9}$ joules every second, into its cooling water at about $300$ K.
Treating the environment as a reservoir, the entropy it gains each second is $Q/T = 2 \times 10^{9}/300 \approx 6.7 \times 10^{6}$ J/K. That entropy has to go somewhere: it is the unavoidable price of extracting work from heat, and the second law says no engine can avoid paying it. Where the cooling water is a river, the Clean Water Act's limits on thermal discharge cap how warm it may leave. The Browns Ferry nuclear plant on the Tennessee River has had to reduce its output during several hot summers, most recently in 2011, because the river was already too warm to accept the heat within its permit.
Raising the temperature at which a plant takes in heat, or lowering the temperature at which it rejects it, shrinks the entropy each joule of work costs — which is why modern combined-cycle gas plants, taking in heat above $1500$ K, reach efficiencies near $60$ percent.
A refrigerator moves heat out of its cold interior, lowering that interior's entropy. The second law forbids the total from falling, so it must dump more entropy somewhere else — and it does, through the warm coils on its back.
Take a kitchen refrigerator that removes $100$ J of heat from its interior at $277$ K ($4$ °C). The interior loses $100/277 = 0.361$ J/K. For the total not to fall, the kitchen at $295$ K must gain at least $0.361$ J/K, which means receiving at least $295 \times 0.361 = 106.5$ J of heat. The extra $6.5$ J is the least electrical work the refrigerator must draw to move the $100$ J.
Real refrigerators create extra entropy in their compressor and through the large temperature differences across their coils, and draw three to five times that minimum. That is why the coils feel warm: they carry away not only the heat taken from the food but the work of the compressor as well, and with it every bit of entropy the whole process created.
Energy and entropy are often learned together, and it is tempting to treat entropy as another conserved quantity that only moves around. It is not. In every irreversible process — heat crossing a temperature difference, friction, mixing, a gas rushing into a vacuum — entropy is created from nothing, and the total rises. Only in an idealized reversible process is the total unchanged. The difference between energy and entropy is exactly the difference between the first law and the second.
A second trap is to believe the Clausius relation applies only to reversible processes and so is useless for real ones. Entropy is a state function: its change depends only on the start and end states. To find the entropy change in an irreversible process, imagine any reversible path between the same states and compute $\int dQ/T$ along it. That is how the mixing example works: neither portion of water was heated reversibly, but each ends in a state that a slow heating would also have reached.
$1.0$ kg of ice melts at $0$ °C. Find the heat it absorbs.
$Q = mL_f = 1.0 \times 333\,000 = 333\,000\ \text{J}$
The latent heat of fusion of ice is $333$ kJ/kg.
Convert the temperature to kelvin.
$T = 0 + 273 = 273\ \text{K}$
Dividing by a Celsius temperature of zero would be meaningless.
Note that the temperature stays fixed.
$\Delta S = \int\dfrac{dQ}{T} = \dfrac{Q}{T}$
Melting happens at a single temperature, so $T$ comes out of the integral.
Divide the heat by the temperature.
$\Delta S = \dfrac{333\,000}{273} \approx 1220\ \text{J/K}$
Joules over kelvin.
Interpret the sign.
$\Delta S > 0$
Liquid water has far more microstates than ice: its molecules are free to move and rotate.
$1.0$ kg of water is heated from $20$ °C to $80$ °C. Write the entropy change for a small step.
$dS = \dfrac{dQ}{T} = \dfrac{mc\,dT}{T}$
Each small amount of heat enters at the current temperature.
Integrate between the two temperatures.
$\Delta S = mc\ln\dfrac{T_f}{T_i}$
The integral of $1/T$ is $\ln T$.
Convert to kelvin.
$T_i = 293\ \text{K}, \qquad T_f = 353\ \text{K}$
A ratio needs absolute temperatures.
Evaluate the logarithm.
$\ln\dfrac{353}{293} = \ln 1.2048 \approx 0.1863$
Using a calculator.
Multiply by the heat capacity.
$\Delta S = 1.0 \times 4186 \times 0.1863 \approx 780\ \text{J/K}$
The entropy the water gains.
Check with an average temperature.
$\dfrac{Q}{T_{\text{avg}}} = \dfrac{4186 \times 60}{323} \approx 778\ \text{J/K}$
For a modest range the heat over the average temperature is close; the logarithm is exact.
$1.0$ kg of water at $80$ °C and $1.0$ kg at $20$ °C mix in an insulated basin. Find the final temperature.
$T_f = \dfrac{80 + 20}{2} = 50\ ^{\circ}\text{C} = 323\ \text{K}$
Equal masses of the same liquid meet halfway.
Find the heat exchanged.
$Q = mc\Delta T = 1.0 \times 4186 \times 30 = 125\,580\ \text{J}$
Each portion changes by $30$ degrees.
Find the hot water's entropy change.
$\Delta S_h = 4186\ln\dfrac{323}{353} = 4186 \times (-0.0888) \approx -372\ \text{J/K}$
It cools, so it loses entropy.
Find the cold water's entropy change.
$\Delta S_c = 4186\ln\dfrac{323}{293} = 4186 \times 0.0975 \approx 408\ \text{J/K}$
It warms by the same amount but gains more entropy, because it started colder.
Add the two changes.
$\Delta S = -372 + 408 = 36\ \text{J/K}$
The basin is insulated, so these are the only changes.
Say what the positive total means.
$\Delta S > 0 \quad\Rightarrow\quad \text{irreversible}$
The water will never separate back into hot and cold by itself.
Find the work that could have been extracted instead.
$W_{\text{lost}} \approx T_0\,\Delta S = 293 \times 36 \approx 1.1 \times 10^{4}\ \text{J}$
An engine running between the two portions could have delivered about $11$ kJ of work; mixing them threw it away.
Compare the lost work with the heat exchanged.
$\dfrac{1.1 \times 10^{4}}{1.26 \times 10^{5}} \approx 8.4\%$
Only a modest fraction of the heat could ever have become work, because the temperature difference is small.
Find the entropy the hot reservoir loses.
$\dfrac{Q}{T_h} = \dfrac{1200}{400} = 3.0\ \text{J/K}$
Heat over its temperature.
Find the entropy the cold reservoir gains.
$\dfrac{Q}{T_c} = \dfrac{1200}{300} = 4.0\ \text{J/K}$
The same heat at a lower temperature.
Subtract the loss from the gain.
A lake at $500$ K absorbs $700$ kJ of heat from a power plant. Its temperature does not measurably change. By how much does its entropy change?
Complete the worked solution: $900$ J flows from a reservoir at $400$ K to one at $200$ K. How much entropy is created?
Divide the heat by the hot temperature.
$\dfrac{Q}{T_h} = \dfrac{900}{400} =$ a J/K
This is the entropy the hot reservoir loses.
Divide the heat by the cold temperature.
$\dfrac{Q}{T_c} = \dfrac{900}{200} =$ b J/K
This is the entropy the cold reservoir gains.
Subtract the loss from the gain.
$\Delta S = \dfrac{Q}{T_c} - \dfrac{Q}{T_h} =$ c J/K
The entropy created by the irreversible flow.
Match each process, for a $1$ kg sample, to its entropy change.
| $Q/T$ | $mL/T$ | $mc\ln(T_f/T_i)$ | $0$ | |
|---|---|---|---|---|
| heat into a reservoir | ||||
| melting or boiling at the transition temperature | ||||
| warming with a constant specific heat | ||||
| slow expansion in an insulated cylinder |
$500$ J of heat leaks through a rod from a reservoir at $400$ K to one at $250$ K. Fill in each reservoir's entropy change and the total, in J/K.
| entropy change (J/K) | |
|---|---|
| hot reservoir | |
| cold reservoir | |
| total |
What is the entropy change of $0.6$ kg of a substance in this phase change: liquid nitrogen boiling at 77 K? Its entropy change per kilogram is $L/T = 2584.4$ J/(kg K). Give the answer in J/K.
Answer: J/K
$0.4$ kg of water is heated from $10$ °C to $60$ °C. What is its change in entropy, in J/K? Use $c = 4186$ J/(kg K).
Answer: J/K
In a kitchen sink, $0.6$ kg of water at $60$ °C is mixed with $0.6$ kg at $20$ °C in an insulated basin. How much entropy is created, in J/K? Use $c = 4186$ J/(kg K).
Answer: J/K
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
$300$ J of heat leaks through a rod from a reservoir at $250$ K to one at $200$ K. Fill in each reservoir's entropy change and the total, in J/K.
| entropy change (J/K) | |
|---|---|
| hot reservoir | |
| cold reservoir | |
| total |
You can find entropy changes from measured heats and temperatures. Explain to someone why heat flowing from a hot body to a cold one always creates entropy, even though no energy is created.
15. Your turn: $1200$ J leaks through a rod from a reservoir at $400$ K to one at $300$ K. How much entropy is created?, step 3
$\Delta S = 4.0 - 3.0 = 1.0\ \text{J/K}$
Positive: the flow from hot to cold creates entropy.