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The efficiency of a heat engine is given by n = Wu / QH Where Wu - is the useful work (7190 J) QH - Heat taken from the hot reservoir (13,8 x 4184 J) 1 cal = 4,184 J => 1 kcal = 1000 cal = 4184J So n = 7190 / (13,8 x 4,184 x 1000) J and since you want that in percent multiply the result by 100.
99.4%
it is the amount of work that can be done. a machine can use 300,000 J of energy but it only uses 263,000 J. that is percent efficiency.
The efficiency of a pulley system depends on how it is set up. Be specific.
Cuz nuffin is purfect
A reservoir contains 75 percent saltwater
Light water nuclear plants like PWR and BWR have efficiency of about 33 percent, that is the ratio of electric output to reactor thermal output. Gas cooled reactors can be up to 40 percent as they work at higher temperature.
Not if you consider the energy expended to grip or engage the lever--this does account for a minimum amount of lost efficiency
The efficiency of a heat engine is given by n = Wu / QH Where Wu - is the useful work (7190 J) QH - Heat taken from the hot reservoir (13,8 x 4184 J) 1 cal = 4,184 J => 1 kcal = 1000 cal = 4184J So n = 7190 / (13,8 x 4,184 x 1000) J and since you want that in percent multiply the result by 100.
nothing has 100% efficiency.
Perhaps an electrical AC transformer can have 99 percent efficiency. A loudspeaker can have only 1 percent efficiency.
The maximum efficiency of the carnot engine only depends on two factors: 1 - The temperature of the hot reservoir (TH) 2 - The temperature of the cold reservoir (TC) And is given by (TH - TC) / TH « or » 1 - TC / TH So by that we can see the maximum efficiency (100%) would be when the difference of temperatures between the two reservoirs is infinite.
99.4%
it is the amount of work that can be done. a machine can use 300,000 J of energy but it only uses 263,000 J. that is percent efficiency.
watches
The laws of thermodynamics imply that there will always be some loss of efficiency.
It is not a good efficiency engine.