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

Asked by Abhisek | 1 year ago |  143

1 Answer

Solution :-

According to law of conservation of energy

Total energy = work done + internal energy

ΔQ = ΔW + ΔU


Rate at which heat is supplied ΔQ = 100 W

Rate at which work is done ΔW = 75 Js-1

Rate of change of internal energy is ΔU

ΔU = ΔQ – ΔW

ΔU = 100 – 75

We get,

ΔU = 25 J s-1


The internal energy of the system is increasing at a rate of 25 W

Answered by Pragya Singh | 1 year ago

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