# How is “the surrounding” and its temperature defined to calculate entropy change during a reaction?

Lets say supercooled liquid water at $$263\ \mathrm K$$ isobarically changes to solid ice at the same temperature. I wish to calculate the change in entropy of the surroundings and I happen to know the $$\Delta H$$ for the reaction.

This should be simple enough. Since the process is isobaric, $$q_\text{rev} = \Delta H$$ and $$q_\text{surr}=-q_\text{reac}$$

$$\Delta S_\text{surr} = \frac{-\Delta H}{T}$$

But what should T be here? Is it $$263\ \mathrm K$$? Or the temperature of the surroundings? Or are they both the same, ie;

Do we consider the temperature at which the reaction is taking place to be the temperature of the surroundings as well?

If so, why?

• The phase transition no reaction. Yes. That would be the minum for that process – Alchimista Jan 19 '19 at 9:41
• Correction: Unless its an adiabatic process the surroundings have to be at the same T as the system (isothermal), otherwise its not at equilibrium. – Buck Thorn Jan 19 '19 at 12:53
• In thermodynamics, the surroundings are typically treated as an ideal constant temperature reservoir whose temperature T is specified. And for such a reservoir, its entropy change is always $q_{surr}/T$, irrespective of any irreversibilities in the system. Also, the temperature at the interface between the system and the surroundings is always the reservoir temperature, not the average temperature of the system. – Chet Miller Jan 19 '19 at 13:30
• You are not giving a realistic problem. The entropy of the surrounding is independent of the system unless you say otherwise. The surrounding will only be affected if there is heat transfer. You haven't said if this is the case, and it is unreasonable to assume either way. – Charlie Crown Feb 19 '19 at 2:47