Question

If a system is isolated the heat transfer between the system and the surroundings is zero i.e DeltaU =0 then how can entropy be a +ve value?

Answer 1

We can explain increase in entropy using Clausius inequality.
We know that $\delta w-\delta w_{rev}\ge 0$
where $\delta w_{rev}$ is the reversible work.
First law of thermodynamics states that
$du=\delta q+\delta w=\delta q_{rev}+\delta w_{rev}$
Now since internal energy u is a state function, all paths between two states (reversible or irreversible) results to same change in u.
Now in an isolated system $du$ = 0.
So, $\delta w-\delta w_{rev}=\delta q_{rev}-\delta q\ge 0$
Hence, $\frac{\delta q_{rev}}{T}\ge \frac{\delta q}{T}$
again we know that entropy change is: $ds=\frac{\delta q_{rev}}{T}$
So, $ds=\frac{\delta q_{rev}}{T}\ge \frac{\delta q}{T}$
For an isolated system : $\delta q=0\Rightarrow ds\ge 0$