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Physics 11 Online
OpenStudy (anonymous):

A gasoline engine receives 200 J of energy from combustion and loses 150 J as heat to the exhaust. What is its efficiency? 75% 25% 33% 125%

OpenStudy (abhisar):

Efficiency = \(\sf 1-\huge \frac{Q_2}{Q_1}\)

OpenStudy (abhisar):

\(\sf Q_2\)= Heat lost and \(\sf Q_1\) = Input heat

OpenStudy (abhisar):

When calculating the efficiency in %, multiply the result with 100

OpenStudy (anonymous):

75%?

OpenStudy (abhisar):

How?

OpenStudy (anonymous):

150/200 = .75 x 100 = 75

OpenStudy (abhisar):

\(\color{blue}{\text{Originally Posted by}}\) @Abhisar Efficiency = \(\sf \huge 1-\frac{Q_2}{Q_1}\) \(\color{blue}{\text{End of Quote}}\)

OpenStudy (anonymous):

Isn't that still 75?

OpenStudy (abhisar):

oops! Sorry..correction..

OpenStudy (abhisar):

\(\sf \frac{200-150}{200}\times 100 =?\)

OpenStudy (abhisar):

Not getting it yet?

OpenStudy (anonymous):

25?

OpenStudy (abhisar):

Yes, that's \(\huge \checkmark\)

OpenStudy (anonymous):

Thanks!

OpenStudy (abhisar):

Welcome :)

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