Heat transfer with surroundings: why considered reversible?

I do not understand how the fact that the surroundings are large makes heat transfer with them reversible? I understand that the temperature of the surroundings is not going to change in the limit, but what about processes in which there is a definite temperature gradient with the system, such as a highly exothermic reaction?

I was sure that this question had already been asked somewhere on Stack, but I can't find anything similar – if you know of one, I would really appreciate if you could refer me to it.

The surroundings are typically modelled in thermodynamics as an ideal (reversible) reservoir with infinite heat capacity and negligible temperature gradients (i.e., infinite thermal conductivity). The assumption is that the temperature gradients are confined to the system, rather than the surroundings, and the entropy generation is confined to the system. Of course, in real life, this is only an approximation.

• Thank you! So for example, if you burn something, do you consider the burning material and the shell of hot air it generates as the "system", and put the boundary between the system and the surroundings where the temperature difference between the air inside the shell and the air outside it becomes infinitesimal? Feb 14, 2017 at 23:19
• Sorry, how do I reference your name? @Chester Miller and any variations aren't recognised. Feb 14, 2017 at 23:20
• Apparently it did. Feb 15, 2017 at 0:52
• No. The temperature difference at the boundary is always infinitesimal. Temperature is continuous at a boundary. I'm not talking about an interface with air outside the shell, but with an ideal reservoir (e.g. a liquid) featuring an infinite heat capacity and infinite thermal conductivity. Feb 15, 2017 at 0:56
• You would have to treat the surroundings as a separate system and you would have to solve the transient partial differential momentum balance and heat transfer equations for the entire process for both the system and surroundings. This would include viscous effects and heat conduction/convection effects. In most cases, it is not possible to apply equilibrium thermodynamics to an irreversible process. Feb 16, 2017 at 13:56