Question

An electric heater supplies heat to a system at a rate of 120 W. If system performs work at a rate of 80 J s–1, the rate of increase in internal energy is

• A. 30 J s–1
• B. 40 J s–1
• C. 50 J s–1
• D. 60J s–1

Answer 1

Answer: (B)

40 J s–1

Answer 2

: According to first law of thermodynamics

$\Delta Q = \Delta U + \Delta W$

$\therefore\frac{\Delta Q}{\Delta t} = \frac{\Delta U}{\Delta t} + \frac{\Delta W}{\Delta t}$

Here , $\frac{\Delta Q}{\Delta t} = 120W,$

$\frac{\Delta W}{\Delta t} = 80Js^{- 1}$

$\therefore\frac{\Delta U}{\Delta t} = 120 - 80 = 40Js^{- 1}$