Question

An electric appliance supplies $6000\, J / \min$ heat to the system. If the system delivers a power of $90 \,W$. How long it would take to increase the internal energy by $2.5 \times 10^{3} J$ ?

• A. $2.5 \times 10^{2} s$
• B. $4,1 \times 10^{1} s$
• C. $2.4 \times 10^{3} s$
• D. $2.5 \times 10^{1} s$

Answer 1

Answer: (A)

$\Delta Q =\Delta U +\Delta W$
$\frac{\Delta Q }{\Delta t }=\frac{\Delta U }{\Delta t }+\frac{\Delta W }{\Delta t }$
$\frac{6000}{60} \frac{ J }{ \sec }=\frac{2.5 \times 10^{3}}{\Delta t }+90$
$\Delta t =250 \,\sec$