wiz-icon
MyQuestionIcon
MyQuestionIcon
2
You visited us 2 times! Enjoying our articles? Unlock Full Access!
Question

A thermally insulated rigid container contains an ideal gas at 27oC. It is fitted with a heating coil of resistance 50Ω. A current is passed through the coil for 10 minutes by connecting it to a d.c. source of 10 V. The change in the internal energy is


A

zero

No worries! We‘ve got your back. Try BYJU‘S free classes today!
B

300 J

No worries! We‘ve got your back. Try BYJU‘S free classes today!
C

600 J

No worries! We‘ve got your back. Try BYJU‘S free classes today!
D

1200 J

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

The correct option is D

1200 J


Heat produced is given by
dQ=V2tR=(10)2×(10×60)50=1200 J
Since the container is rigid, the change in volume dV = 0. Hence work done dQ = PdV = 0. From the first law of thermodynamics, the change in internal energy is dU = dQ - dW = dQ = 1200 J.


flag
Suggest Corrections
thumbs-up
1
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Power Delivered and Heat Dissipated in a Circuit
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon