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In thermodynamics, whenever heat is released from the system to the surroundings, temperature of surroundings remains unchanged but entropy increases. But entropy is a function of temperature. How is it possible?

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  • $\begingroup$ can you mention the source which says that says "whenever heat is released from the system to the sorrounding , temperature of sorrounding remains unchanged but entropy increases" you have got it wrong whenever trempreature passes from someting more hotter to something less hotter there is a increasein trempreature of the less hotter substance as well as the trempreature of the less hotter substance $\endgroup$ – user24065 Dec 24 '15 at 6:29
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In scenarios like the one you describe, the surroundings are typically modeled as an infinite reservoir, with the capacity to supply or absorb heat with negligible change in temperature. If you treated it as a large finite reservoir instead, and evaluated its entropy change based on $\Delta S=mc\ln ((T_0+\Delta T)/T_0)$ and its heat absorbed by $Q=mc\Delta T$ (where m is the mass of the reservoir and c is its heat capacity), you would find that, in the mathematical limit of "mc" becoming infinite (at constant Q), the entropy change would approach $\Delta S=Q/T_0$.

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The entropy function is of the form $S(T,p,n) = n\, s(T,p)$, where $s(T,P)$ is the entropy per mole.

The surrounding is always much, much larger than the system. Hence, for the surrounding, $n$ is an extremely large number. Then, $S(T,p,n)$ can change by a finite amount with little change in $s(T,p)$, in which case $T$ and $p$ also stay more or less the same.

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Practically speaking, although the external temperature may not appear to increase (e.g. the room doesn't get noticeably hotter) the energy is transferred to the kinetic energy of the surroundings and the average energy of the surroundings increases fractionally. This increases the entropy of the surroundings. Depending on the the circumstances that energy can be used to do work, i.e. increasing the volume of the universe which also corresponds to an increase in entropy.

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Start with entropy function $S=S(T,V,N)$. Then

$$dS = \frac{\partial S}{\partial T}dT + \frac{\partial S}{\partial V}dV + \frac{\partial S}{\partial N}dN$$

Since entropy is an extensive quantity and temperature is an intensive quantity, it follows that the coefficient $(\partial S/\partial T)$ has to be extensive as well. The bigger is the system (e.g. the environment) the bigger is this coefficient and then smaller is the variation of temperature. If volume and composition are held constant and if the change in entropy is of order $dS \sim 10$ in arbitrary units and $(\partial S/\partial T) \sim 10^3$, then $dT \sim 10^{-2}$.

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