Let’s analyze the situation step by step:

System: The liquid inside a thermally insulated (adiabatic) and closed vessel.

Process: The liquid is mechanically stirred from outside.

Heat Exchange ($q$)

Because the vessel is thermally insulated, there is no heat exchange between the system and surroundings. Hence,

$ q = 0. $

Work ($w$)

The external mechanical stirring does work on the system by agitating the liquid. Work done on the system is conventionally taken as positive:

$ w > 0. $

Change in Internal Energy ($\Delta U$)

From the First Law of Thermodynamics,

$ \Delta U \;=\; q \;+\; w. $

Since $q = 0$ and $w > 0$, we get

$ \Delta U = w > 0. $

Hence:

$ \boxed{ \Delta U > 0,\quad q = 0,\quad w > 0. } $

Answer: Option B is correct.