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Question:
An electric appliance supplies $6000 \mathrm{~J} / \mathrm{min}$ heat to the system. If the system delivers a power of $90 \mathrm{~W}$. How long it would take to increase the internal energy by $2.5 \times 10^{3} \mathrm{~J}$ ?
Correct Option: 1
Solution:
$\Delta \mathrm{Q}=\Delta \mathrm{U}+\Delta \mathrm{W}$
$\frac{\Delta \mathrm{Q}}{\Delta \mathrm{t}}=\frac{\Delta \mathrm{U}}{\Delta \mathrm{t}}+\frac{\Delta \mathrm{W}}{\Delta \mathrm{t}}$
$\frac{6000}{60} \frac{\mathrm{J}}{\mathrm{sec}}=\frac{2.5 \times 10^{3}}{\Delta \mathrm{t}}+90$
$\Delta \mathrm{t}=250 \mathrm{sec}$
Option (1)
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