Physics

Grade-9

Easy

Question

# Your heating system is 55 percent energy-efficient. What amount of energy would it consume to transform 9000 kWh into useful thermal energy for heating the house during the winter?

- 16363.63 kWh
- 163.63 kWh
- 1.6363 kWh
- 1636 kWh

Hint:

### Efficiency (%) = (useful energy output ÷ energy input) × 100

## The correct answer is: 16363.63 kWh

- Efficiency (%) = (useful energy output ÷ energy input) × 100

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