# Do irreversibilites change the final temperatures of two bodies?

When two identical bodies are brought into thermal contact, and both are surrounded by an isolation, does the fact that they have a finite temperature difference, which generates entropy, change their final temperatures. Are their final temperatures their arithmetic average

• By the way in general case the specific heat is a function of temperature so in general case final temperature is not the arithmetic average – Aleksey Druggist Jul 26 at 12:19
• You commented the following regarding my original and subsequently deleted answer "I read that irreversible heat transfer is due to entropy generation less useful according to the area under the T-S diagram where a rev. heat transfer corresponds to a larger area than the irrev. one. As justification there was the sentence that heat is a path function and its value depends one the path taken. So I concluded that irreversible heat transfer decreases the actual heat transferred and therefore the temperature does not rise like in a rev. HT. Process" – Bob D Aug 2 at 16:26
• I realized that I didn’t fully understand what you read not seeing it in context. Are you able to edit your question to cite the source and proved the context of what your read? – Bob D Aug 2 at 16:26
• @AlekseyDruggist I think this is an answer. Why not post it like that? – Claudio Saspinski Aug 2 at 18:08
• @ClaudioSaspinski I don't think AlekseyDruggist comment is an answer. I believe the OP is assuming the heat capacity is constant over the specified temperature range and wants to know whether or not entropy generation alters the final temperature, which it does not, and why – Bob D Aug 2 at 19:44

Let's consider a bathtub full of hot water @ $$T_1 = 35°C$$ and a "small" stone @ temperature $$T_2 = 5°C$$. From this setup is should be clear that if we bring the two "bodies" together, the final temperature will not be their arithmetic average, $$\bar T = (T_1 + T_2)/2 = 20°C$$, but that it is much closer to $$35°C$$.