Question

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?

Answer 1

Heat energy supplied per second by the heater ( ∆Q ) = 100 w
Workdone by the system ( ∆W) = + 75 j/s
Rate of change of internal energy ( ∆U) = ?

A/c to 1st law of thermodynamics,
∆Q = ∆W + ∆U
100 = 75 + ∆U
∆U = 100 - 75 = 25 J/s
Internal energy increasing 25 joule per second .