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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 \(2.5 \times 10^{3}\, {J}\) ?

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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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