JEE Mains · Physics · STD 11 - 11. thermodynamics
A source supplies heat to a system at the rate of \(1000 \,W\). If the system performs work at a rate of \(200\,W\). The rate at which internal energy of the system increases \(.......\,W\)
- A \(1200\)
- B \(600\)
- C \(500\)
- D \(800\)
Answer & Solution
Correct Answer
(D) \(800\)
Step-by-step Solution
Detailed explanation
\(dQ=dU+dw\) \(\frac{\mathrm{dU}}{\mathrm{dt}}=\frac{\mathrm{dQ}}{\mathrm{dt}}-\frac{\mathrm{dw}}{\mathrm{dt}}\) \(\frac{\mathrm{dU}}{\mathrm{dt}}=1000-200=800 \mathrm{~W}\)
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