So, I know that the internal energy follows the equation:

$dU = dQ + dW$

and that for a constant pressure enthalpy follows the equation:

$dH = dU + P dV$.

But since the work done on the system is just $dW = -P dV$ (work being done on the system compresses it), $H = Q$.

So what's the difference between internal energy and heat? Are they equal or is work not just caused by compression?

  • $\begingroup$ Since $dH=dU+pdV$ if $\delta W = -pdV$ then $dH=\delta Q$ and not as you have written $H=U$ $\endgroup$
    – hyportnex
    Mar 29, 2017 at 19:10
  • $\begingroup$ @hyportnex: Sorry, that's what I meant. I made a mistake while typing the question. $\endgroup$
    – abc123
    Mar 29, 2017 at 19:13
  • $\begingroup$ In general $H \ne Q$ even if $dH = \delta Q $. After integrating you can only get the rather misleading equality $H=Q + h_0$ for some unknown integration constant $h_0$ if you assume that all external work is reversible and against constant pressure $\delta W = -pdV$. $\endgroup$
    – hyportnex
    Mar 29, 2017 at 19:26

2 Answers 2


It's not entirely wrong to think of enthalpy as the heat content of a system1, but due to the definition of heat as a certain type of energy transferred between systems, there are subtle differences.

Enthalpy is a state function, a priori only well-defined after equilibrium has been established, whereas heat is a characteristic of the specific process $p$ that takes you from initial state $i$ to final state $f$.

Let $$ \Delta H_{i\to f} = H(f) - H(i) $$

In general, we have $$ \Delta H_{i\to f} = Q(p) + \Delta H^\text{uncomp}(p) $$ where we can have an additional change in enthalpy due to entropy production within the system (eg when friction is involved) that is not compensated by entropy loss of the environment due to heat transfer.

For quasistatic processes where state variables are well-defined at all times, we also have $$ \Delta H_{i\to f} = \int_p T \mathrm dS $$

In that case, we may go to an infinitesial description $$ \mathrm dH = T\mathrm dS = \delta Q + T \mathrm \delta S^\text{uncomp} $$ where the $\delta$ forms need only be defined along the specific curve through phase space.

In case of reversible processes, we additionally have $$ \delta S^\text{uncomp} = 0 $$ and thus $$ \mathrm dH = \delta Q \qquad\qquad \Delta H_{i\to f} = Q(p) $$ which was your starting point.

1 or rather, as Wikipedia puts it, "the capacity to do non-mechanical work plus the capacity to release heat"


dU = dQ + dW

dH = dU + PdV =dQ + dW + PdV = dQ ( from third given equation)

How did you get H = U at first place?


dH = dU + d(PV)

Yes H = Q but only if pressure is constant


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