# Proving that Entropy change of the system become zero for quasistatic process

Suppose a body with initial temperature to be, say $$T_0$$ and it cooled down to temperature $$T_f$$ (say) by contacting it to series of heat baths. I want to show that in the limit when heat baths become infinite, so the change in the system's entropy gets zero.

What I have done is the following first for a body that is at temperature $$T_s$$ and final temperature (when contacted with heat bath with temperature $$T_r$$) $$T_r$$ the change in the entropy of the system is turn out to be

$$\Delta S_{sys}=\Delta S_{body}+\Delta S_{bath}$$ $$\Delta S_{sys}=C\ln\left(\frac{T_r}{T_s}\right)+C\left(\frac{T_s-T_r}{T_r}\right)$$ Now suppose we change do the series change like $$T_0\rightarrow T_0+\Delta T\rightarrow T_0+2\Delta T\cdots T_0+(N-1)\Delta T\rightarrow T_0+N\Delta T=T_f$$

So that the change in the system's entropy is $$\Delta S_{sys}=C\ln\left(\frac{T_f}{T_0}\right)-C\left( \frac{\Delta T}{T_0+\Delta T}+\frac{\Delta T}{T_0+2\Delta T}+\cdots +\frac{\Delta T}{T_0+N\Delta T}\right)$$

I don't know what to do next. If I need to get zero that means that $$\lim_{N\rightarrow \infty}\left[ \frac{\Delta T}{T_0+\Delta T}+\frac{\Delta T}{T_0+2\Delta T}+\cdots +\frac{\Delta T}{T_0+N\Delta T}\right]=\ln\left(\frac{T_f}{T_0}\right)$$

But I don't How this suppose to come. And If my steps are correct or not?

• It's probably just me, but I'm having trouble following this. Let's start with what it is that you are defining as "the system". It seems you are defining the system as the combination of the body and the bath. Is that correct? If so, then what is the system's surroundings? Commented Dec 26, 2020 at 21:36
• this may help you physics.stackexchange.com/questions/317690/… Commented Dec 26, 2020 at 22:44
• @BobD Yes! I'm considering the system consists of the bath and the body. They are isolated from the rest of the universe and so surrounding. Commented Dec 27, 2020 at 4:55

Your summation approaches $$\int_{T_i}^{T_f}{\frac{dT}{T}}$$ as N becomes infinite. The summation should really be $$\sum_{I=1}^{N}\frac{(T_f-T_i)/N}{T_i+(T_f-T_i)i/N}$$
• There is no difference, You have just taken $\Delta N=(T_f-T_i)/N$. How this supposed to help me prove that entropy change is zero. Commented Dec 27, 2020 at 4:57
You are trying to prove a false statement. In the reversible process (i.e. quasi-static without friction) you can write the form of heat transfer as: $$$$\Delta Q = T dS.$$$$ Therefore, in a reversible process, an increase in heat causes an increase in entropy.
I am not sure if you are wrong. Where did the expression $$$$\Delta 𝑆_{𝑠𝑦𝑠}=𝐶 \ln\left(\frac{𝑇_𝑟}{𝑇_𝑠}\right)+𝐶\left(\frac{𝑇_𝑠−𝑇_𝑟}{𝑇_𝑟}\right)$$$$ come from?