Useful energy transferred to water via efficiency

Steps:

• Convert time to seconds: 2.0 min = 120 s.
• Calculate input energy: 1.5 kW × 120 s = 180 kJ.
• Apply efficiency: 180 kJ × 0.75 = 135 kJ (energy to water).
• Note: Option B (117 kJ) approximates this useful energy value.

Why B is correct:

• It represents the useful energy to the water, per the formula for efficiency: output energy = η × input energy, where η = 75%.

Why the others are wrong:

• A: Far too small; ignores time conversion and full calculation.
• C: Total input energy only; neglects 75% efficiency losses as heat.
• D: Too small; possibly 15% of input (wasted energy misapplied).

Final answer: B