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ASmith:
Why is the heat flow from a system divided by T equated to the entropy change in the surroundings?  Some of this heat could be stored in other systems and so would not appear as entropy.  (To be pedantic it's the magnitudes of H_sys/T and S_surr that are equated as there is a difference in sign.)

ASmith:
As no one has replied yet perhaps I should explain a bit more.  I recently read a chemistry textbook (which I found more interesting than expected) and it says ΔS_univ = ΔS_sys +ΔS_surr = ΔS_sys -ΔH_sys/T.  This is where the equality is assumed.  After multiplying by T we get TΔS_sys -ΔH = G = Gibbs free energy.

Suppose an internal combustion engine is ticking over and doing no work.  All of the heat is radiated into the surroundings and can be said to contribute to entropy.  But suppose some of the heat from the exhaust is used (via the Seebeck effect for example) to store energy in a battery.  This heat wouldn't contribute to the entropy of the surroundings.  So the equality isn't correct.  Where am I going wrong?

Orcio_Dojek:

--- Quote ---Suppose an internal combustion engine is ticking over and doing no work.  All of the heat is radiated into the surroundings and can be said to contribute to entropy.  But suppose some of the heat from the exhaust is used (via the Seebeck effect for example) to store energy in a battery.  This heat wouldn't contribute to the entropy of the surroundings.  So the equality isn't correct.  Where am I going wrong?
--- End quote ---
ΔS_univ = ΔS_sys +ΔS_surr

Enthropy of the system (engine) will not change, as all the heat will be radiated out.

Some of the heat can be used to store energy in a battery, but never 100 %, so enthropy of the surrounding will still rise.

ASmith:
Thanks for replying. (By enthropy I assume you mean entropy, not enthalpy.)  The entropy of the surroundings seems to rise by different amounts in the two different situations, but the heat emitted by burning a given amount of fuel will be the same, i.e. ΔH.  The problem is getting the equality to work in different situations.

Suppose when no energy is stored, the surroundings are at 25 deg.C and the heat flow, ΔH, increases the entropy of the surroundings by ΔS.  In the second case suppose 20% of the fuel's energy is stored in a battery using an alternator.

If the engine (i.e. the system) is kept at the same temperature in both cases and the Gibbs energy stays the same then TΔS is constant (from G = ΔH - TΔS.)   But only 80% of the heat now contributes to the entropy of the surroundings.  So to keep TΔS the same, the surroundings need to be 25% hotter.  By my reckoning this means the temperature needs to rise to about 100 deg.C.

Assuming this doesn't happen, and also assuming that the Gibbs equation gives the right answers, it seems obvious that it's the conversion of chemical energy into enthalpy, ΔH, which is important in determining the viability of reactions as far as the surroundings are concerned.  This is the quantity used in the equation.  The change of entropy per se is irrelevant.

Orcio_Dojek:
Engine (or a car on the whole) loses enthalpy, as it burns out the fuel (quantity of fuel is source of its potential energy -> enthalpy).

Its entropy will not change, as - even if it generates out the heat it is still the same engine (car) as before.

As of surroundings....

Some of the heat can be stored back in a battery, rest is dispersed in the air.

Heat stored in the battery is calculated as increase of the enthalpy, while dispersed in the air - as increase of the entropy of the surrounding.

After all....

ΔG = ΔG_car + ΔG_surroundings

ΔG = (ΔH - TΔS)_car + (ΔH - TΔS)_surroundings

If some of the heat is stored as an energy in a battery term ΔH_surr will increase at cost of TΔS_surr.

But ΔG for whole process should not change.