# Entropy change in a calorimetry problem

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 $dQ = mcdT$?

• infinitesimal "heat" exchange over almost the same temperature is reversible: $TdS = \delta Q = mcdT$ so then $dS = \frac {mc}{T}dT$ – hyportnex Feb 14 '16 at 0:23
• 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. – Chet Miller Feb 14 '16 at 2:35

Step 3: Calculate the integral of dQ/T for the reversible path you have identified. This is $\Delta S$
$$\Delta S\geq \int (dQ/T)_B$$where B signifies the boundary where the heat transfer is occurring.