Question

A rigid container with thermally insulated walls contains a gas and a coil of resistance 50 \(\Omega\), carrying a current of 1 A. The change in internal energy of the gas after 2 minutes will be

• A. 6 kJ
• B. 10 kJ
• C. 3 kJ
• D. 12 kJ
• E. 1.5 kJ

Hint:

Whenever you see "rigid container," assume no mechanical work is done by or on the gas. All energy input directly affects internal energy.

Answer 1

The Correct Option is A

Concept: This problem applies the First Law of Thermodynamics and Joule's Law of heating.
• First Law: \(\Delta U = Q - W\). In a rigid container, volume is constant (\(dV = 0\)), so work done \(W = 0\).
• Insulated Walls: No heat enters or leaves the system from outside, but internal heating via a coil adds energy (\(Q_{in} = H\)).
• Joule's Heating: Heat produced \(H = I^2 Rt\).

Step 1: Calculate the total heat generated.
Convert time to seconds: \(t = 2 \text{ min} = 120 \text{ s}\). Given \(I = 1 \text{ A}\) and \(R = 50\ \Omega\): \[ H = (1)^2 \times 50 \times 120 = 6000 \text{ Joules} \]

Step 2: Determine change in internal energy.
Since \(W = 0\), all the heat generated by the coil increases the internal energy: \[ \Delta U = 6000 \text{ J} = 6 \text{ kJ} \]