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Thermodynamics question

2019 · 9 Apr · Shift 2 · Q16
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Thermodynamics question

2019 · 9 Apr · Shift 2 · Q16

JEE MainChemistryThermodynamicsMCQ+4 / −1
During compression of a spring the work done is 10kJ and 2kJ escaped to the surroundings as heat. The change in internal energy, Δ\DeltaΔ U(inkJ) is :
  1. A
    - 12
  2. B
    8
  3. C
    - 8
  4. D
    12
View written solutionFree

Correct answer: B

  1. Use the first law of thermodynamics

    ΔU=q+w\Delta U = q + wΔU=q+w

    where:

    • qqq = heat absorbed by the system
    • www = work done on the system

  2. Identify the signs

    • During compression of the spring, work is done on the system: w=+10 kJw = +10\,\text{kJ}w=+10kJ

    • 2 kJ2\,\text{kJ}2kJ heat escaped to the surroundings, so the system lost heat: q=−2 kJq = -2\,\text{kJ}q=−2kJ

  3. Substitute into the first law

    ΔU=(−2)+10=8 kJ\Delta U = (-2) + 10 = 8\,\text{kJ}ΔU=(−2)+10=8kJ

  4. Match with the options

    ΔU=8 kJ\Delta U = 8\,\text{kJ}ΔU=8kJ

    So the correct option is B.

  5. Comparison with stored answer

    Stored correct answer: B

    Derived answer: B

    They agree.

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