First slide

Heat ;work and internal energy

Question

A rigid container with thermally insulated walls contains a coil resistance $100 Ω$ carrying current 1 A change in internal energy after 5 minutes will be

Moderate

A

0 KJ
B

20 KJ
C

10 KJ
D

30 KJ

Solution

work done(dW) = heat energy produced in the resistor
$dW=dU$ =change in internal energy
$\begin{array}{rcl} {heat energy produced in a resistor is dW=i}^{2} Rt \\ g i v e n c u r r e n t i & = & 1 A; r e s i s \tan c e R = 100 Ω; t i m e t = 5 x 60 = 300 s \\ s u b s t i t u t e g i v e n v a l u e s i n a b o v e e q u a t i o n \\ heat energy produced in a resistor & = & 1 \times 100 \times 300 \\ \Rightarrow & c h a n g e i n i n t e r n a l e n e r g y = 30000 J \\ \Rightarrow & c h a n g e i n i n t e r n a l e n e r g y = 30 kJ \end{array}$

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