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The total change in entropy is given by $$\mathrm dS_\text{total}=\mathrm dS_\mathrm{system}+\mathrm dS_\mathrm{surroundings}$$ Now, the entropy change for the system is $$\mathrm dS_\mathrm{system}=\mathrm … dS_\mathrm{produced}+\mathrm dS_\mathrm{exchanged}$$ where, $\mathrm dS_\mathrm{produced}$ is internally produced entropy in the system and $\mathrm dS_\mathrm{exchanged}$ is entropy change due to exchange …

If $Q_\mathrm h$ amount of heat flow from source to sink, then change in entropy of source is $\frac{-Q_\mathrm h}{T_\mathrm h}$ and change in entropy of sink is $\frac{+Q_\mathrm h}{T_\mathrm l}$. … And while I understand that entropy is being generated in the partition, my question is how does heat conduction across a finite temperature gradient generates entropy …

When one of the weights is reduced( for expansion), the pressure on the piston drops suddenly and the piston shoots up, so there will be temperature gradients inside the gas, which will give rise to entropy …

Entropy change of a system is a state function... … Shouldn't the entropy of the system in both cases must be same? This answer: How would one calculate the entropy change for an adiabatic irreversible process? …

I know how to calculate entropy change of the system but am not able to calculate entropy change for surroundings. … I was able to calculate the entropy change of system. It came out to be, $\Delta S_{sys} = 123JK^{-1}mol^{-1}$. But I am stuck at change in entropy of surroundings. …