Question

An electric heater supplies heat to a systematic at rate of 100w. If Systems performs work at rate of 75 joules per second. At what rate is the internal energy increasing ?​

Answer 1

Heat is supplied to the system at a rate of 100 W.

∴Heat supplied, Q = 100 J/s

The system performs at a rate of 75 J/s.

∴Work done, W = 75 J/s

From the first law of thermodynamics, we have:

Q = U + W

Where,

U = Internal energy

∴U = Q – W

= 100 – 75

= 25 J/s

= 25 W

Therefore, the internal energy of the given electric heater increases at a rate of 25 W.

Answer 2

$<marquee behaviour=move bgcolor=green><h1><strong><u>⤵</u></strong>Here is Your Answer<strong><u>⤵</u></strong><h1></marquee>$

☞Heat supplied (Q)= 100J per sec

☞Rate of performance of system = 75 J/sec

☞Work done (W)= 75 J/sec

☞According to first law of thermodynamics....

☞Q =U + W

☞U = Q - W

☞= 100 -75

☞=25 W

✔25 W is the answer✔