Out of the 200KJ input, 150KJ isn't usable. The only usable output of the
machine is the remaining 50KJ.
The efficiency of that machine is 50KJ/200KJ = 25% .
It must be pointed out, though, that it all depends on your definition of what is
usable and non-usable output. For example, if this machine were labeled as an
"Office Space Heater", then the 150KJ of heat would be the usable output, and
the efficiency would have been 75% .
25 percent
25 percent
25%
25 percent
If the waste heat is 15 compared to the total heat input of 20 (ignoring the thousands) this means the heat doing useful work is 5, which represents 25 percent of the total input heat, so this is the efficiency. Waste heat in a car is the sum of the engine cooling losses and the exhaust losses, the rest is converted to mechanical energy which is useful. 25%
Efficiency is measured in joules.
200,000-150,000= 50,000 25 %
400
The efficiency is 80%. To find the efficiency, 400/500 = 80%.
The efficiency is 80%. To find the efficiency, 400/500 = 80%.
If the input energy is 210 joules and the efficiency of the system is 30%,then the output energy is30% of 210 = (0.3 x 210) = 63 joules.
The machine efficiency is 35 percent (35/100).