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JEE Mains · Physics · STD 11 - Thermodynamics
- JEE Mains 2021
- Medium
- Numerical
An electric appliance supplies \(6000\, {J} / {min}\) heat to the system. If the system delivers a power of \(90\, {W}\). How long (in \(sec\)) it would take to increase the internal energy by \(2.5 \times 10^{3}\, {J}\) ?
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(250\)
Step-by-step Solution
Detailed explanation
\(\Delta \mathrm{Q}=\Delta \mathrm{U}+\Delta \mathrm{W}\)
\(\frac{\Delta \mathrm{Q}}{\Delta \mathrm{t}}=\frac{\Delta \mathrm{U}}{\Delta \mathrm{t}}+\frac{\Delta \mathrm{W}}{\Delta \mathrm{t}}\)
\(\frac{6000}{60} \frac{\mathrm{J}}{\mathrm{sec}}=\frac{2.5 \times 10^{3}}{\Delta \mathrm{t}}+90\)
\(\Delta \mathrm{t}=250 \,\mathrm{sec}\)
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