Since this heater is 60% efficient, this means that the heater is losing 40% of the energy. If the actual energy required is 56.5 J, then 0.4 of 56.5 is being lost and will be needed additionally by the heater to heat the water to its final temperature. 0.4×56.5 = 22.6 kJ Hence, the total energy required is 22.6+56.5 kJ = 79.1 kJ
It takes 56.5 kilojoules of energy to raise the temperature of 150 milliliters of water from 5°C to 95°C. If you use an electric water heater that is 60% efficient, how many kilojoules of electrical energy will the heater actually use by the time the water reaches its final temperature?