Specific Heat Solids, Liquids, and Gases Clarification/Confirmation

Question

$Q = mcDelta T$ (1)
where
$Q = mbox{heat}$
$m = mbox{mass}$
$c = mbox{specific heat}$
$Delta T = mbox{change in temperature}e$
$Q = c_vnDelta T$ (2)
where
$Q = mbox{heat}$
$n = mbox{number of moles}$
$c_v = mbox{specific molar heat at constant volume}$
$Delta T = mbox{change in temperature}e$
(1) is used to find the heat transfer for solids and liquids. (2) is used to find the heat transfer for gas molecules at constant volume(I am aware there is one for constant pressure). Can (2) be converted to (1) by looking up the molecules molar mass in the periodic table and converting the number of moles to grams? If that is possible, is the reason why we do not use $Q = mcDelta T$ for gas is because the specific heat depends on the type of process?

Chet Miller · Accepted Answer

The real problem here is associating any of this with the heat transferred Q.  In thermodynamics, Q is not a function of state (but rather process), while C is a function of state, not process.  Because, in thermodynamics, there can be work done on the system, the amount of heat changes depending on how much work was done.  And, different amounts of heat can be associated with the same change from a specified initial state to a specified final state.
All these issues and ambiguities are cleared up in thermodynamics by defining heat capacity, not in terms of heat Q, but in terms of the thermodynamics state functions internal energy (per unit mass or mole) and enthalpy (per unit mass or mole):
$$C_V=left(frac{partial U}{partial T}right)_V$$
$$C_P=left(frac{partial H}{partial T}right)_P$$
The nice thing is that these more precise definitions apply to all phases: gas, liquid, and solid.

Specific Heat Solids, Liquids, and Gases Clarification/Confirmation

One Answer

Add your own answers!

Ask a Question