Question

A rigid container with thermally insulated walls contains a coil of resistance 100 $\Omega$, carrying current 1 A. Change in internal energy after 5 min will be

• A. Zero
• B. 10 kJ
• C. 20 kJ
• D. 30 kJ

Answer 1

The Correct Option is D

W = 0. Therefore, from first law of thermodynamics, 
\(\, \, \, \, \, \, \Delta U = \Delta Q = i^2Rt\)
\(\, \, \, \, \, \, \, \, \, \, = (I)^2 (100) (5 \times 60)J = 30kJ\)
\(\therefore\) Correct answer is (d)

Answer 2

We are given a coil with a resistance of 100 Ω carrying a current of 1 A. The power dissipated by the coil is determined by the formula:

P = I² R

Substituting the given values:

P = (1 A)² × 100 Ω = 100 W

The energy dissipated (which in this thermally insulated system increases the internal energy) over a time period t is:

ΔU = P × t

Here, t is given as 5 minutes. Converting minutes to seconds:

t = 5 min × 60 s/min = 300 s

Therefore, the change in internal energy is:

ΔU = 100 W × 300 s = 30,000 J

Change in Internal Energy = 30,000 J (or 30 kJ)

Since the container is rigid and thermally insulated, all the electrical energy dissipated is retained within the system, increasing its internal energy.