# I Conditions for writing Q=CΔT

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1. Aug 18, 2016

### FG_313

When dealing with heat transfer, there are cases where Q can be expressed by C*ΔT, for some proportionality constant C. However, in isothermal processes for example, this formula would lead to a mistake, for any value of C (because it would imply Q=0, wich is not true generally speaking). What about a general process going from an initial state to a final one? Is it always possible to find such C?
Edit: I`ve just realised that I wrote "righting" instead of "writing" in the title. Sorry. <Moderator's note: title edited>

Last edited by a moderator: Aug 22, 2016
2. Aug 18, 2016

### Staff: Mentor

In freshman physics, we learned that, when heat is added to a constant volume system, we can write Q = CΔT, where C is called the heat capacity. However, when we got more deeply into the basics and learned thermodynamics, we found that this elementary approach is no longer adequate (or precise). We found that Q depends on process path and that, if work W is occurring, this changes things. However, we still wanted C to continue to represent a physical property of the material being processed, and not to depend on process path or whether work is occurring. This is dealt with in thermodynamics by changing the definition of C a little. Rather than associating C with the path dependent heat Q, in thermodynamics, we associate C with parameters relating to the state of the material being processed, in particular internal energy U and enthalpy H. We define the heat capacity at constant volume $C_v$ as the derivative of the internal energy U with respect to temperature at constant volume:
$$C_v=\left(\frac{\partial U}{\partial T}\right)_v\tag{1}$$
We also found that we could define a heat capacity at constant pressure $C_p$ as the derivative of the enthalpy H with resepct to temperature at constant pressure:$$C_p=\left(\frac{\partial H}{\partial T}\right)_p\tag{2}$$
The question is, "do either of these definitions reduce to the more elementary version from freshman physics under any circumstances." The answer is "yes." From the first law of thermodynamics, we find that, for a closed system of constant volume (no work being done), $Q=\Delta U=C_v\Delta T$, and, for a closed system experiencing a constant pressure change (with $W=p\Delta v$), $Q=\Delta H=C_p\Delta T$. Of course, Eqns. 1 and 2 are much more generally applicable than this.

3. Aug 21, 2016

### FG_313

Great answer, thank you very much!