an engine does 1500J of useful work with each 5000J of energy supplied to it.
what is its efficiency? what happens to the rest of the energy supplied?
2007-05-22 07:24:04 · 6 answers · asked by jessica 1 in Science & Mathematics ➔ Physics
1500J/5000J = 30% efficiency
The rest is lost as waste heat.
2007-05-22 07:29:53 · answer #1 · answered by feanor 7 · 0⤊ 0⤋
Efficiency = Useful Output/Total Input = 1500/5000 =.3 or 30%.
Without analyzing the engine, we can only speculate where the 3500 J of potential work disappeared to. However, typical losses are through the exhaust, where energy leaves as heat. Other losses are wherever parts move and useful energy is lost again as heat...friction heat.
Finally, some of that input energy is often used to drive auxiliary equipment; so if "useful" is defined as energy expended in performing the primary function of the engine (e.g., to power a vehicle drive train), then the energy going to the auxiliaries (e.g., an air conditioner) is also lost.
2007-05-22 14:40:34 · answer #2 · answered by oldprof 7 · 0⤊ 0⤋
Efficiency is measured as production divided by input.
As to what happens to the rest of the energy, think about your car's engine. Have you ever put your hand on a car's hood after it's been running a while?
2007-05-22 14:30:27 · answer #3 · answered by Brian L 7 · 0⤊ 0⤋
the efficiency is 1500/5000 * 100. That is the percent energy that is used (efficiency). the rest of the energy is mostly heat.
2007-05-22 14:30:24 · answer #4 · answered by mr.quark 2 · 0⤊ 0⤋
Efficiency = 0.3 or 30%
Rest of energy is waste, heat and overcoming friction.
2007-05-22 14:35:46 · answer #5 · answered by Anonymous · 0⤊ 0⤋
efficiency 1500/5000=0.3 =30%
The rest is spend to heat
2007-05-22 14:29:33 · answer #6 · answered by maussy 7 · 0⤊ 0⤋