Here's the equation:

$\Delta S = \frac{\Delta q}{T}$

The confusion I have with this equation lies in the variable for temperature, $T$. Why is the temperature constant in this equation? Why can't the equation include $\Delta T$ instead of just $T$ and, surely, if $\Delta q$ is negative or positive the temperature must decrease or increase, respectively.

  • 2
    $\begingroup$ The temperature is not constant in this equation. This equation is just a simplification for cases when the temperature change is negligible and you don't need to take care of it. All the other cases, you should do the integration. $\endgroup$
    – Greg
    Jun 21, 2015 at 23:16
  • $\begingroup$ So, say the temperature change is not negligible, could one replace T with ΔT, or does a significant temperature change involve a much more complex equation? $\endgroup$
    – kmcmillan
    Jun 22, 2015 at 0:42
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    $\begingroup$ The equation for $\Delta S$ will be very different depending on the conditions; however, all are derived from the infinitesimal relation in Shadock's answer, which happens to be the definition of entropy itself. Most commonly if temperature changes but pressure is kept constant one would use $\mathrm{d}q = C_p\,\mathrm{d}T$ and integrate over the temperature range to get $\Delta S = C_p\ln\frac{T_2}{T_1}$ assuming $C_p$ is constant over the temperature range. $\endgroup$ Jun 22, 2015 at 2:17
  • $\begingroup$ I would like to add one comment that by the Second Law of Thermodynamics, the infinitesimal equation only holds true if the heat transfer is reversible: $\mathrm{d}S = \frac{ \mathrm{d}q_{\mathrm{rev}}}{T}$. However, since $S$ is a state function, even if the entire process is not reversible, this has no effect on the value of $\Delta S$ since $\Delta S$ only depends on the initial and final states and not the process by which the system goes from the initial to the final states. $\endgroup$ Jun 22, 2015 at 2:19
  • $\begingroup$ @kmcmillan You keep T, and you integrate the expression with this T changing along the process. $\endgroup$
    – Greg
    Jun 22, 2015 at 4:59

1 Answer 1


Your relation is not correct, so you must not understand something. If I were you I will asked the question in Physics SE but nevermind.

The good relation is : $$\mathrm{d}S=\frac{\delta Q}{T}$$

I let Richard Feynman explain it to you here (when you click on the link wait few seconds)

I hope it will help you, have fun, live long and do chemistry !

  • 1
    $\begingroup$ $dS=\frac{\delta q}{T}$ is just the infinitesimal version of $\Delta S =\frac{\Delta q}{T}$ and still implies that $T$ is constant or minimally changing in comparison to $q$. The OP is interested in knowing how this equation would be different if $T$ varied significantly (or perhaps that $q=Q(T)$. $\endgroup$
    – Ben Norris
    Jun 22, 2015 at 2:09
  • $\begingroup$ And there is a big difference ... $\endgroup$
    – ParaH2
    Jun 22, 2015 at 2:10
  • $\begingroup$ Agreed, but it does not answer the OP's question (especially if the OP is only at the general chemistry level). It would also be helpful to summarize the content of your link to combat link rot. $\endgroup$
    – Ben Norris
    Jun 22, 2015 at 2:13
  • $\begingroup$ I am french and don't speak english as well as i understand it so if you are you will do better than me :) $\endgroup$
    – ParaH2
    Jun 22, 2015 at 2:18
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    $\begingroup$ @BenNorris No, it doesnt imply that T is constant. To obtain the whole change of S, you have to integrate the infinitesimal expression, where T can vary at every point. $\endgroup$
    – Greg
    Jun 22, 2015 at 4:57

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