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An electric appliance supplies

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 $2.5 \times 10^{3} \mathrm{~J}$ ?

  1. $2.5 \times 10^{2} \mathrm{~s}$

  2. $4.1 \times 10^{1} \mathrm{~s}$

  3. $2.4 \times 10^{3} \mathrm{~s}$

  4. $2.5 \times 10^{1} \mathrm{~s}$

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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