Thank you for reporting, we will resolve it shortly
Q.
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:
BHUBHU 2006
Solution:
$W=0$ Therefore, from first law of thermodynamics
$\Delta U=\Delta Q=i^{2} R t$
$=(1)^{2} \times 100(5 \times 60) $
$=30\, k J$