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. $60 \,J \,s^{-1}$

Answer 1

Answer: (B)

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