1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Confusion with internal energy and enthalpy

  1. Sep 24, 2013 #1
    For incompressible substances you have a change in internal energy represented by

    ΔU = mcΔT

    where T is temperature.

    I've also seen

    ΔH = mcΔT and ΔQ = mcΔT

    where H and Q are enthalpy and heat, respectfully.

    Does that mean that all of these are the same? I heard H becomes U for open systems...

    I'm so confused about this. Thanks
  2. jcsd
  3. Sep 24, 2013 #2


    User Avatar
    Staff Emeritus
    Science Advisor

    The basic relationship is

    ΔU = ΔQ - pΔV
    where p is pressure and V is volume.

    We can rearrange this as:

    ΔQ = ΔU + pΔV

    We can interpret this as the statement: When you add heat (ΔQ) to the system, the system can respond by increases its internal energy (ΔU) or by increasing its volume. So if the system expands, it's hard to keep track of how much energy has been added to the system.

    If we switch to enthalpy, H, it is defined by:

    H = U + PV

    So ΔH = ΔU + pΔV + VΔp

    Therefore, we can re-express the equation for ΔQ as follows:

    ΔQ = ΔH - VΔp

    Now, in common laboratory experiments, the pressure is kept constant. So Δp = 0. In such a case,

    ΔQ = ΔH

    But that equation is only true when pressure is constant.
  4. Feb 13, 2015 #3
    Quick follow-up on this topic: I know that, for a closed, constant V system, if we want to find the Q exchanged in say a heating process, we solve Q=integral(Cv*dT) over the range of temperatures. However, if you have a constant P system, then you solve Q=integral(Cp*dT) over the T's. My question is why doesn't this calculation also work for the closed, constant V system? It just seems that if you applied the enthalpy approach to the const V system, then the pressure term should subtract out and you would just be left with the same answer as using the internal energy approach. But I'm pretty sure the answers are different.
  5. Feb 13, 2015 #4
    For a constant volume heating process, ##Q = ΔU=\int{C_vdT}##
    For a quasistatic constant pressure heating process, ##Q = ΔH=\int{C_pdT}=ΔU+PΔV##,
    The Q's in these equations are different. So, in the latter situation, ##ΔU=ΔH-PΔV=Q-PΔV##
    If you are dealing with ideal gases, then the enthalpy approach works for the constant V system because, for ideal gases, U and H are functions only of T. So, for example, for constant volume, ##Q = ΔU=\int{C_vdT}=ΔH-VΔP=\int{C_pdT}-VΔP##. But for an ideal gas at constant volume:
    ##VΔP=RΔT##. So, ##\int{C_vdT}=\int{(C_p-R)dT}##. This is consistent with the relationship between Cv and Cp for an ideal gas: ##C_v=C_p+R##.

    For real gases beyond the ideal gas region, both U and H are functions not only of temperature but also of pressure (or volume). So this approach breaks down. However, it is important to remember the more general definitions of these two heat capacities (not involving Q):

    $$C_v=\left(\frac{\partial U}{\partial T}\right)_V$$
    $$C_p=\left(\frac{\partial H}{\partial T}\right)_P$$

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook