# Irreversible Adiabatic Process [closed]

While studying about the entropy change of surroundings in irreversible adiabatic process, I came across a sentence.

For infinite reservoir, all processes are considered reversible

And suddenly it resumes with the equation:

q/T = dS (surrounding)

Can anyone explain the line and how the conclusion was drawn?

• Can you provide a reference? What are you using as a convention for the sign of q? – Buck Thorn Oct 8 '19 at 20:56
• I am using the sign convention as in chemistry. Anything that increases the total energy is considered positive. But my real confusion is that sentence and it's connection with the next line formula. – Hardik Oct 9 '19 at 5:22
• Are you copying the equation verbatim? If using the convention that q>0 when heat is transferred from the system to the surroundings, then the equation makes sense. – Buck Thorn Oct 9 '19 at 10:47
• q is the heat of the process. From the second law we know I know that q(rev)/T= dS. But to describe it for the surroundings only, the limitation reversible is removed. This is what I thought, but it just confuses me over. – Hardik Oct 9 '19 at 11:40
• Possible duplicate of Calculating entropy change of surroundings – Buck Thorn Oct 9 '19 at 12:33

You should also keep in mind that heat is a signed quantity, and that there can be different conventions with regard to what is a positive/negative heat transfer. Usually chemists say q>0 when the system receives energy as heat, and when the surroundings loses energy as heat. Since heat is regarded as the sole source of change in the entropy of the surroundings (assumed to be an inifinitely large reservoir at constant T$$^\ast$$), we write $$dS_{surroundings} = -\frac{dq}{T}$$ for an infinitesimal heat transfer and $$\Delta S_{surroundings} = -\frac{q}{T}$$ for a finite heat transfer.
$$^\ast$$ An infinite reservoir can provide or accept energy in the form of heat without temperature change (or any other change in its state).