A standard textbook problem has us calculate the change in entropy in a system that undergoes some sort of heat exchange. For example, object $A$ has specific heat $c_a$ and initial temperature $T_A$ and object $B$ has specific heat $c_b$ with initial temperature $T_B$. They are they put in contact with each other until they reach thermal equilibrium, and our goal is to find the total entropy change of the system.

The standard solution is to use $$S = \int \frac{dQ}{T}$$ where $dQ = mcdT$. But the above integral is only satisfied for reversible processes, whereas this heat exchange is clearly irreversible.

The usual workaround for this is to pick some reversible path and calculate the entropy change on our "fake" path, since entropy is a state variable. For example, in the free expansion of an ideal gas, we pick calculate the entropy change along an isotherm that carries us along the expansion to find the true change in entropy.

My question is - what exactly is the reversible path we are using when we use $dQ = mcdT$?

  • $\begingroup$ infinitesimal "heat" exchange over almost the same temperature is reversible: $TdS = \delta Q = mcdT$ so then $dS = \frac {mc}{T}dT$ $\endgroup$
    – hyportnex
    Feb 14, 2016 at 0:23
  • $\begingroup$ As I said in my answer below, one simple reversible path involves gradually contacting each body with a continuous sequence of constant temperature baths running from the original temperature of the object to its final equilibrium temperature. $\endgroup$ Feb 14, 2016 at 2:35

1 Answer 1


In calculating the entropy change for an irreversible process between two thermodynamic equilibrium states of a system, the first step in the procedure should always be to FORGET ALL ABOUT THE IRREVERSIBLE PROCESS PATH ENTIRELY. The entropy change can only be calculated for a reversible path between the same two end states.

Step 2: Identify a convenient reversible process path between the same two thermodynamic equilibrium states. Any reversible path will do, because they all give the same result for the entropy change.

Step 3: Calculate the integral of dQ/T for the reversible path you have identified. This is $\Delta S$

For the example that you gave, I like use a process path where I start out by separating the two objects entirely, and then contacting each of them with a continuous sequence of constant temperature baths running from the original temperature of the object to its final equilibrium temperature. This gives me the integral that you alluded to in your question for each body.

For an irreversible process path, the temperature of the system is not even uniform spatially throughout, so what temperature do you use in evaluating the integral of dQ/T? Clausius noted that, if you use the temperature at the boundary where the heat transfer is occurring (say for one of the bodies), you always find that the integral you get over the irreversible path is less than the entropy change of the body. This is called the Clausius inequality:

$$\Delta S\geq \int (dQ/T)_B$$where B signifies the boundary where the heat transfer is occurring.


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