Question

An electric heater supplies heat to a system at rate of 100 W. If system performs works 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.

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Answer 2

$\bold{Answer}$

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

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