A liquid when kept inside a thermally insulated closed vessel at 25°C was mechanically stirred from outside. What will be the correct option for the following thermodynamic parameters ? (1) triangle of U > 0, q = 0, w > 0 (2) triangle of U = 0, q = 0, w = 0 (3) triangle of U < 0, q = 0, w > 0 (4) triangle of U = 0, q < 0, w > 0

Key Concepts

1. First Law of Thermodynamics
   $\Delta U = q + w$
   Where
   • $\Delta U$ = change in internal energy of the system
   • $q$ = heat exchanged with surroundings (positive if absorbed)
   • $w$ = work done on the system (positive in chemistry sign-convention)

2. Thermally Insulated (Adiabatic) Vessel
   An ideal thermal insulator means no heat can flow:
   $q = 0$

3. Closed System
   Mass does not cross the boundary, but energy can (as work).

4. Mechanical Stirring
   The paddle/stirrer pushes the fluid. In chemistry sign convention, the surroundings (your hand/motor) do work on the system, so
   $w > 0$

Logical Chain of Thought

• Start with the information “thermally insulated” → immediately set $q = 0$.
• Recognise that stirring is a form of shaft work → work is supplied → $w$ is positive.
• Apply the First Law:
  $\Delta U = 0 + w$
  Since $w$ is positive, $\Delta U$ must be positive.
• Compare with the given answer choices. Only the choice that has $\Delta U > 0,\; q = 0,\; w > 0$ fits.

Hence, option 1 is correct.

Imagine a closed metal bottle full of water.

• The bottle is wrapped with a perfect thermal blanket, so no heat can enter or leave.
• You put a spoon through a small airtight hole and stir the water very hard.
• Because you are pushing the water around, you are doing work on the water – just like rubbing your hands warms them up.
• Since no heat can escape and you are adding energy by stirring, the water’s own energy (we call it internal energy) must go up.

So:

1. Heat given or taken ($q$) = 0 (blanket stops heat)
2. Work done on the water ($w$) = positive (your muscles add energy)
3. Change in internal energy ($\Delta U$) = positive (because $\Delta U = q + w$)

That matches option 1.

Step-by-Step Solution

1. Identify heat flow
   The vessel is thermally insulated:
   $q = 0$

2. Recognise type of work
   Mechanical stirring (shaft work) is performed on the system:
   $w > 0$

3. Apply First Law
   $\Delta U = q + w$ $\Delta U = 0 + w$ Since $w$ is positive,
   $\Delta U > 0$

4. Match with options Only option (1) states

   • $\Delta U > 0$
   • $q = 0$
   • $w > 0$

[ \boxed{\text{Option (1) is correct}} ]