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

Explanation

Heat is supplied to the system at a rate of one second dQ = 100 J
The work done by the system per second dW=75 J
Increase in internal energy per second `DeltaU`=?
From the first law of thermodynamics
`dQ=dU+dW`
`therefore` dU=dQ-dW
=100-75
`=25 "Js"^(-1)`
Therefore, the internal energy dU of the given electric heater increases at a rate of 25 W.