One can define the heat capacity of isobaric processes as $$ c_P = \left( \frac{\partial H}{\partial T} \right)_P . $$ Now, we know that the unit of heat capacity is Joule per Kelvin, i.e., I need to put a certain amount of energy into something to heat it by $x$ degrees. According to Wikipedia, for nearly every system we normally look at, heat capacity is positive (which makes sense from our daily interaction with the world). In my professors notes, there's an exercise which asks (verbatim): ''Show a connection between $c_P$ and the derivative of enthalpy and deduce from the curvature of this derivative the sign of heat capacity''.

How can I make a general assumption about enthalpy which yields the positive sign I instinctively assume?


1 Answer 1


We know the relationship between enthalpy, energy, pressure and volume $$H = U + P V \, . \tag{1}$$ Differentiation of this expression yields $$ dH = T dS + V dP \, .$$ However, $$TdS = dQ$$ where dQ is the amount of heat system received or gave away. Take the derivative of equation $(1)$ with respect to $T$ at constant $P$ to get $$\left( \frac{dH}{dT} \right)_p = \left( \frac{dQ}{dT} \right)_p = C_p \, .$$ If $T$ of the system increases then it received heat and $dT$, $dQ$ and $(dQ/dT)_p$ are all positive. Oppositely, if $T$ decreases then $dT<0$ and $dQ < 0$ but $(dQ/dT)_p$ is still positive. Roughly speaking this means that heat capacity is always positive.

  • 1
    $\begingroup$ Welcome to Physics Stack Exchange. Please note that this site supports mathjax. I edited this answer to use it. You can look at how it works by hitting the "edit" button. $\endgroup$
    – DanielSank
    Commented Sep 15, 2015 at 21:59
  • $\begingroup$ Thanks, that makes sense. I guess my professor isn't looking for a depper answer than "differentials are less than 0, derivative is still bigger than zero -> heat capacity still positive". $\endgroup$
    – John Doe
    Commented Sep 16, 2015 at 14:58
  • $\begingroup$ @Nikolay Frik you answer says nothing about curvature, which is the second derivative of "something". $\endgroup$
    – hyportnex
    Commented Sep 17, 2015 at 13:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.