- #1
- 6
- 0
I'm given a fuel cell that uses methane as fuel:
(CH4) + 2(O2) --> 2(H2O) + (CO2)
And I'm asked to find (a) the electrical work I can get out for each mole of methane, and (b) the amount of waste heat produced.
I understand that ΔG is defined as the amount of useful non-mechanical work that can be obtained from the system. So my first thought was that ΔG = Q + W(electrical). However, I'm told it's actually the case that ΔG = W(electrical) only. So why exactly is that? Isn't Q a form of non-mechanical work? Secondly, when I'm asked to compute the amount of waste heat, I go ahead and accept the confusing notion that ΔG = W(electrical) and attempt to compute the waste heat simply as ΔH = ΔU + PΔV = ΔU - W = Q. However, I'm told the waste heat is actually Q = ΔG - ΔH. Can anyone explain these two areas of confusion for me?
(CH4) + 2(O2) --> 2(H2O) + (CO2)
And I'm asked to find (a) the electrical work I can get out for each mole of methane, and (b) the amount of waste heat produced.
I understand that ΔG is defined as the amount of useful non-mechanical work that can be obtained from the system. So my first thought was that ΔG = Q + W(electrical). However, I'm told it's actually the case that ΔG = W(electrical) only. So why exactly is that? Isn't Q a form of non-mechanical work? Secondly, when I'm asked to compute the amount of waste heat, I go ahead and accept the confusing notion that ΔG = W(electrical) and attempt to compute the waste heat simply as ΔH = ΔU + PΔV = ΔU - W = Q. However, I'm told the waste heat is actually Q = ΔG - ΔH. Can anyone explain these two areas of confusion for me?