By Clausius's definition $$dS_{surr} = \frac{\delta q_{rev}}{T}$$ but does this mean $$dS_{surr} = \frac{\delta q}{T}$$ where $\delta q$ is the infinitesimal heat absorbed by the surrounding in any process (may be reversible or irreversible for the system) because the surrounding is always at equilibrium?
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In typical developments, constant temperature reservoirs (that make up part of the surroundings) are treated as ideal reversible entities that receive heat at constant temperature and feature no internal irrversibility. This is just an idealization that can be approached in the real world. The reservoir is assumed to have infinite capacity to absorb heat without its temperature changing, and it is assumed to have very high thermal conductivity so that, for a given heat flux, entropy generation from temperature gradients is minimized.
answered Jan 24, 2016 at 0:33
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$\begingroup$ Ok got it. Thank you! I really need to make sure I'm clear about this point. Since if this is true then I can deduce the equations for $\Delta A$ and $\Delta G$ directly from the equation $dS_{universe}\geq 0$. $\endgroup$– snsunxCommented Jan 24, 2016 at 2:11