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Practice Problems: Applying the First Law of Thermodynamics

Problem 1: No Work Done

Given:
Heat added: Q = +600 J
No work is done: W = 0 J

Step-by-step:

  1. Write the formula: ΔU = Q – W
  2. Substitute the values: ΔU = 600 J – 0 J
  3. Solve: ΔU = 600 J
The internal energy increases by 600 J.

Problem 2: System Does More Work Than Heat Added

Given:
Heat added: Q = 300 J
Work done by system: W = 400 J

  1. Formula: ΔU = Q – W
  2. Substitute: ΔU = 300 J – 400 J
  3. Solve: ΔU = -100 J
The system loses 100 J of internal energy.

Problem 3: Heat Removed, Work Done On the System

Given:
Q = -250 J, W = -150 J

  1. Formula: ΔU = Q – W
  2. Substitute: ΔU = -250 J – (-150 J)
  3. ΔU = -250 J + 150 J = -100 J
Internal energy still drops by 100 J.

Problem 4: Cooling a Gas

Given:
Q = -400 J, W = 100 J

  1. ΔU = Q – W
  2. ΔU = -400 J – 100 J
  3. ΔU = -500 J
The gas's internal energy decreases by 500 J.

Problem 5: Work Done on Gas

Given:
W = -300 J, Q = 0 J

ΔU = Q – W = 0 – (-300 J) = 300 J

The internal energy increases by 300 J.