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

<p>Let’s analyze the situation step by step:</p> <p><p><strong>System</strong>: The liquid inside a <strong>thermally insulated</strong> (adiabatic) and <strong>closed</strong> vessel. </p></p> <p><p><strong>Process</strong>: The liquid is <strong>mechanically stirred</strong> from outside.</p></p> <hr /> <h3>Heat Exchange ($q$)</h3> <p><p>Because the vessel is <strong>thermally insulated</strong>, there is <strong>no heat exchange</strong> between the system and surroundings. Hence, </p> <p>$ q = 0. $</p></p> <h3>Work ($w$)</h3> <p><p>The external mechanical stirring does <strong>work on the system</strong> by agitating the liquid. Work done <strong>on</strong> the system is conventionally taken as <strong>positive</strong>: </p> <p>$ w > 0. $</p></p> <h3>Change in Internal Energy ($\Delta U$)</h3> <p><p>From the First Law of Thermodynamics, </p> <p>$ \Delta U \;=\; q \;+\; w. $</p></p> <p><p>Since $q = 0$ and $w > 0$, we get </p> <p>$ \Delta U = w > 0. $</p></p> <p>Hence:</p> <p>$ \boxed{ \Delta U > 0,\quad q = 0,\quad w > 0. } $</p> <hr /> <p><strong>Answer: Option B</strong> is correct.</p>