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Thermodynamics question

2021 · 25 Jul · Shift 2 · Q16
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Thermodynamics question

2021 · 25 Jul · Shift 2 · Q16

JEE MainChemistryThermodynamicsNumerical+4 / −1
A system does 200 J of work and at the same time absorbs 150 J of heat. The magnitude of the change in internal energy is ‾\underline{\hspace{2cm}}​ J. (Nearest integer)
Numerical answer
View written solutionFree

Correct answer: 50

  1. Use the first law of thermodynamics:

ΔU=q+w\Delta U = q + wΔU=q+w

Here,

  • qqq = heat absorbed by the system = +150 J+150\,\text{J}+150J
  • www = work done on the system

Since the system does 200 J200\,\text{J}200J of work, work is done by the system, so

w=−200 Jw = -200\,\text{J}w=−200J

  1. Substitute into the formula:

ΔU=150+(−200)=−50 J\Delta U = 150 + (-200) = -50\,\text{J}ΔU=150+(−200)=−50J

  1. The question asks for the magnitude of change in internal energy:

∣ΔU∣=50 J|\Delta U| = 50\,\text{J}∣ΔU∣=50J

Therefore, the required nearest integer is:

50\boxed{50}50​

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