JEE Main · 2019 · Shift-IIeasyTHERMO-049

During compression of a spring the work done is 10 kJ and 2 kJ escaped to the surroundings as heat. The change in…

Thermodynamics · Class 11 · JEE Main Previous Year Question

Question

During compression of a spring the work done is 10 kJ and 2 kJ escaped to the surroundings as heat. The change in internal energy, ΔU\Delta U (kJ) is:

Options
  1. a

    −8

  2. b

    8

  3. c

    12

  4. d

    −12

Correct Answerb

8

Detailed Solution

🧠 First Law: ΔU=q+w\Delta U = q + w (IUPAC) — work done on spring stays in the system Work is done on the spring (system gains energy). Some of that energy escapes as heat (system loses energy). Net change = what stays.

🗺️ Calculation w=+10 kJw = +10\text{ kJ} (work done on system — positive in IUPAC) q=2 kJq = -2\text{ kJ} (heat escaped from system — negative)

ΔU=q+w=2+10=+8 kJ\Delta U = q + w = -2 + 10 = +8\text{ kJ}

The system gained a net 8 kJ of internal energy.

⚠️ Trap: "Heat escaped" means q<0q < 0 for the system. Don't add +2 kJ — it left the system, so it's −2 kJ.

Answer: (b)\boxed{\text{Answer: (b)}}

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