Let's say we have a block of lead (of heat capacity $C$) at temperature $T$, and a heat bath of temperature $T_0$. The block is placed in the bath and it cools to $T_0$ (a clearly irreversible process). How can we go about working out the entropy change of both the block itself and of the surroundings (i.e. the heat bath)?

First of all, work doesn't make sense in this context and so it's clear that the block is only in thermal contact with the heat bath (so only heat can be exchanged). The first law of thermodynamics tells us that:

$$dU=dQ+dW=TdS-pdV$$ $$dU=dQ=TdS$$

From this we have a differential expression for $dS$, and armed with the heat capacity of the block, we can perform the integration to find $\Delta S$ for the block, and we can follow a similar method for the bath.

There are two issues I have with this:

  1. How can $dU=dQ$ when the former is a total differential (because $U$ is a function of state) and the latter isn't ($Q$ is not a function of state and depends on path)?
  2. This leads us to $dS=\frac{dQ}{T}$. However, we know from Clausius' theorem that $dS>\frac{dQ}{T}$, unless $dQ=dQ_{rev}$, i.e. the process is reversible, which in this case it clearly isn't.
  • $\begingroup$ It is probably easier to calculate the difference between the initial and the final entropies of the block, is the entropy is a state function. $\endgroup$
    – Roger V.
    Apr 7, 2020 at 12:47
  • $\begingroup$ What alternate reversible process are you using to evaluate the change in entropy of the block? In terms of the initial and final temperatures of the block, what does this give you for its change in entropy? $\endgroup$ Apr 7, 2020 at 12:47
  • $\begingroup$ Note that Clausius' theorem says that $dS_{sys} > \delta Q / T_{surr}$. $\endgroup$
    – J. Murray
    Apr 7, 2020 at 14:26

1 Answer 1


Just calculate $\Delta S=mc \ln(T_2/T_1)$ for the block, assuming it's an incompressible solid (we'll derive this below). Entropy is a state function so the process of how you aquired this change doesn't matter. Now to answer your questions:

  1. This is true because the path/process determines $Q$, which changes the state of the system, hence changing $U$.

  2. You are correct, this is irreversible if you just drop a hot rock in cold water. But you could drop the rock in many infinitesimally cooler baths, thus cooling it reversibly, and $dS=dQ/T$ would apply for each of these increments. Now integrate over all these infinitesimally small coolings, with $dQ=mc\ dT$, and get $\Delta S=mc \ln(T_2/T_1)$. Entropy is a state function, so we don't care about the path we took to get from initial to final state.


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