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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% .

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Q: What is the efficiency of an engine when the heat input is 200000 joules and the waste heat is 150000joules?
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