# Difference between reversible and irreversible heat transfer

Reference books define reversible heat transfer as heat transfer occurs across a infinitesimal thermal gradient i.e. dT and irreversible heat transfer as heat transfer across finite thermal gradient i.e. ΔT.

Now i want to calculate the quantity $$\int _1^2 \frac{dq}{T}$$ for a process involving heat transfer into a system where its temperature changed from $$T_{1}$$ to $$T_{2}$$. My question is if i calculate this integral for a reversible and and an irreversible heat transfer into a system whether they will be equal?

If you do it right, the integrals for a reversible path and for the irreversible path will not be equal. According to Fermi (Thermodynamics) and Moran et al (Fundamentals of Engineering Thermodynamics), the temperature that must be used in these integrals should be that at the interface between the system and the surroundings through which the heat dq is flowing. In a reversible heat transfer process, where the system temperature is virtually spatially uniform throughout, the boundary between the system and surroundings is at essentially the same temperature as the system. However, in an irreversible heat transfer process, if heat is flowing across the boundary into the system, the boundary temperature will be higher than that of the system. This will mean that, for the same total amount of heat transferred, the integral for the irreversible process will be less than the integral (between the same two end states) for the reversible process.

• Thank you for the answer. Now I am able to digest the second law of thermodynamics also. As per your answer $$\int _1^2 \frac{dq_{rev}}{T} = \int _1^2 \frac{dq}{T} + \sigma$$. $\sigma$ a positive integer. You had given method to find out the integral on the RHS. – Anoop A K Apr 18 at 6:32
• $\sigma$ is definitely not an integer. It is just a positive number. – Chet Miller Apr 18 at 12:07