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?

  1. 16363.63 kWh
  2. 163.63 kWh
  3. 1.6363 kWh
  4. 1636 kWh

hintHint:

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

The correct answer is: 16363.63 kWh


    • Efficiency (%) = (useful energy output ÷ energy input) × 100
    • O u t p u t space E n e r g y space equals fraction numerator I n p u t space over denominator E f f i c i e n c y end fraction cross times 100 equals 9000 over 55 cross times 100 equals 16363.63 space k W h

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