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JEE Main 2021 (26 Aug, Shift 1) · Heat, Work & First Law of Thermodynamics

Heat, Work & First Law of Thermodynamics question from JEE Main 2021 (26 Aug, Shift 1)

Q14JEE Main 2021 · 26 AUG · SHIFT 1Medium

Heat, Work & First Law of Thermodynamics

An electric appliance supplies 6000 J/min heat to the system. If the system delivers a power of 90W. How long it would take to increase the internal energy by 2.5 ×\times 103 J?

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Answer: (a)

  1. Heat input rate: 600060=100 W\frac{6000}{60} = 100\text{ W}; work output rate =90 W= 90\text{ W}
  2. dUdt=100−90=10 W\frac{dU}{dt} = 100 - 90 = 10\text{ W}
  3. t=2.5×10310=250 s=2.5×102 st = \frac{2.5\times10^{3}}{10} = 250\text{ s} = 2.5\times10^{2}\text{ s}
  4. So the answer is (a).
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