Entropy as a State Function
Entropy is a state function. Find its change by adding small reversible heat transfers divided by their temperatures along a path between states. This sum is zero around a closed reversible loop. An isolated system's total entropy cannot decrease.
What a learner can do afterwards
- Computes the entropy change for a reversible isothermal process and for a heated solid
- Shows that a complete cycle has zero net entropy change and explains heat/T cancellation around a reversible cycle
- Finds the entropy change of the surroundings and argues that the total cannot fall
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Entropy S is a state function: its change depends on the initial and final equilibrium states. Choose a reversible path and add each small heat transfer divided by its temperature in kelvin. At constant temperature, delta S = Qrev/T. Heat absorbed is positive; heat released is negative. For an irreversible process, use a reversible path between the same states. Its actual heat divided by temperature need not give its entropy change.
Let 20 g of ice melt at 273.15 K, taking 6680 J from a reservoir held at 293.15 K. The water stays at the melting temperature. A reversible melting path gives an ice entropy gain of 6680/273.15, about 24.46 J/K. The reservoir loses 6680/293.15, about 22.79 J/K. Their sum rises by about 1.67 J/K. The bars compare gain and loss magnitudes; the loss enters the sum with a minus sign.
Any complete cycle restores the working substance's initial state, so its net entropy change is zero, even if the cycle is irreversible. For a reversible cycle, the sum of reversible heat/T is also zero. Follow one reversible route from A to B, then reverse another route to return to A. The zero loop sum makes the two forward-route sums equal. This establishes path independence.
For a solid with constant heat capacity C and no phase change, a small reversible temperature rise dT needs heat C dT. Adding C dT/T gives delta S = C ln(Tf/Ti). Ti and Tf are initial and final temperatures in kelvin; ln means natural logarithm. Heating gives a positive result and cooling a negative one. If heat capacity varies, add the small heat/T contributions instead of treating C as constant.
The second law applies to an isolated system, including all affected surroundings. Add the entropy changes of every part. The total is zero for a reversible process and positive for an irreversible process; it cannot be negative. One part may lose entropy if the other parts gain at least as much. A reservoir held at temperature T has entropy change Q/T, with Q positive into that reservoir.
Calculate entropy change along a reversible path. A cycle has zero net change. For an isolated whole, including all affected surroundings, total entropy cannot decrease.
The teaching and question repairs were written for this dataset. OpenStax entropy material was consulted for facts; its prose and figures are not reproduced.
The replacement readings were written for this dataset after factual study of OpenStax entropy material. The retained video references point to Khan Academy. Video terms have not been verified.
More to read
A reversible path gives the entropy change
Entropy is a property of a thermodynamic state. Its change between two equilibrium states does not depend on the process that connected them. An irreversible process and a reversible process with the same endpoints therefore give the system the same entropy change.
The reversible process provides a calculation route. Divide each small heat transfer by the temperature at which it occurs, then add the contributions. Use kelvin and count heat into the system as positive. At constant temperature, this becomes delta S = Qrev/T.
For example, absorbing 72 J reversibly at a constant 360 K raises entropy by 72/360 = 0.20 J/K. When temperature varies, dividing the total heat by the final temperature is not the general calculation.
For a solid with constant heat capacity C and no phase change, reversible heat for a small temperature rise is C dT. Adding C dT/T gives delta S = C ln(Tf/Ti). The endpoint temperatures must be in kelvin. This also gives the change after irreversible warming between those same states.
Zero heat can accompany an entropy increase
An ideal gas can spread into an evacuated part of a rigid, insulated container. No heat crosses the wall, and the gas does no external work. Its internal energy stays constant. For an ideal gas, its final equilibrium temperature is also unchanged.
This free expansion is irreversible and increases the gas's entropy. Its actual heat transfer is zero, but the entropy formula requires reversible heat. A reversible isothermal expansion between the same endpoints would absorb heat and give the positive entropy change. Substituting the actual zero heat would answer a different question.
Melting at a substance's melting temperature gives another useful comparison path. Reversible melting absorbs heat and raises entropy by Qrev/T. Reversible freezing gives the opposite change.
If a warmer reservoir supplies the melting heat Q, calculate each part separately. The melting material gains Q/Tmelt; the reservoir loses Q/Treservoir. Their sum is positive because Treservoir is greater than Tmelt. Energy transfer balances, but entropy changes need not cancel.
The working substance and the surroundings
A working substance completing a cycle returns to its initial entropy. This holds even for an irreversible cycle, because entropy is a state function. It does not imply that the surroundings return to their initial states.
For a reversible Carnot cycle, the hot isotherm gives the working substance an entropy gain Qh/Th. The cold isotherm gives a loss Qc/Tc. Each reversible adiabatic leg has zero entropy change. Since Qh/Th = Qc/Tc, the changes cancel. The hot reservoir's entropy loss also equals the cold reservoir's gain.
For an irreversible engine, the working substance can still have zero net entropy change over a cycle. The engine and its affected surroundings together produce entropy. Apply the second law to the isolated whole: its total entropy cannot decrease.
Entropy is measured in J/K. It is not an amount of energy that has disappeared. Energy remains conserved while irreversibility can reduce the work obtainable under specified environmental conditions.
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