Question

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

• A. \(\Delta U>0, q=0, w>0\)
• B. \(\Delta U=0, q=0, w=0\)
• C. \(\Delta U<0, q=0, w>0\)
• D. \(\Delta U=0, q<0, w>0\)

Hint:

In a thermally insulated system, no heat exchange occurs (q=0). Work done on the system is positive (w>0), and work done by the system is negative (w\>0).

Answer 1

The Correct Option is A

To solve the given problem, we need to analyze the changes in the thermodynamic parameters of the system when a liquid in a thermally insulated closed vessel is mechanically stirred. The key steps involve understanding the implications of each listed parameter under the given conditions:

1. System Description: The liquid is in a thermally insulated closed vessel, which implies that there is no heat exchange with the surroundings, i.e., q = 0.

2. Work Done: Mechanical stirring involves external work being done on the system. Thus, w > 0, as work is added to the system from the surroundings.

3. Change in Internal Energy (∆U): According to the first law of thermodynamics: \(\Delta U = q + w\). Substituting the known values, we get:

\(\Delta U = 0 + w \gt 0\)

Since w > 0, the internal energy of the system increases, making ∆U > 0.

Therefore, the correct option for the thermodynamic parameters is:

\(\Delta U>0, q=0, w>0\)

Answer 2

Step 1 — Heat exchange: 
The vessel is thermally insulated, so no heat is exchanged with the surroundings: $$q=0.$$ 
Step 2 — Work done:
Mechanical stirring does work on the liquid (energy is supplied to the system), so the work done on the system is positive: $$w>0.$$ 
Step 3 — First law of thermodynamics:
The first law gives $$\Delta U = q + w.$$ Substituting $q=0$ and $w>0$: $$\Delta U = w > 0.$$ 
Conclusion:
$$\boxed{\Delta U>0,\; q=0,\; w>0}$$ Correct option: (1)