Answer:
20000J or 20kJ
Step-by-step explanation:
Given parameters:
Efficiency of the battery = 80%
Amount of energy transferred = 16000J
Unknown:
How much energy was transferred to the battery = ?
Solution:
Since the battery is 80% efficient, it loses 20% of energy anytime it wants to store energy.
Now, 16000J of energy is contained in the battery, the original amount of energy sent to the battery;
Let the amount of energy sent to the battery be y
80% of y = 16000
0.8y = 16000
y = 20000J or 20kJ