A liquid when kept inside a thermally insulated closed vessel $$25^{o}C$$ at was mechanically stirred from outside. What will be the correct option for the following thermodynamic parameters ?

A liquid in a thermally insulated closed vessel is mechanically stirred from outside. We need to find the thermodynamic parameters.

The vessel is thermally insulated, so there is no heat exchange with the surroundings: $$q = 0$$. Mechanical stirring does work on the liquid, and since work done on the system is positive by IUPAC convention, $$w \gt 0$$.

According to the first law of thermodynamics, $$\Delta U = q + w$$. Because $$q = 0$$ and $$w \gt 0$$, we have $$\Delta U = 0 + w \gt 0$$, hence $$\Delta U \gt 0, \; q = 0, \; w \gt 0$$.

The correct answer is Option 3: $$\Delta U \gt 0, q = 0, w \gt 0$$.