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An engine burns 1.4 kg of gasoline per hour and yields 3.5 kW of power. Calculate the efficiency of the engine when the heating value of gasoline is 43 MJ/kg.

Mathematics
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i dont even know where to start
ANY help, even a formula or a rule or a hint, anything will help
i just know efficiency = input/output

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Other answers:

Yep, calm down.,,
oh, sorry, it is output/input
ok,.... i am calm
The heating value is the amount of heat energy produced by complete combustion in a given quantity. Take a basis of 1 hour. Then your engine burns 1.4kg of gasoline, which yields a maximal amount of thermal energy to use for work as\[43MJ/kg \times 1.4 kg=60.2MJ\]In one hour, your engine produces from this energy, \[\frac{3500J}{s}\times \frac{3600s}{h}=12.6MJ\]The efficiency is the amount of work per unit energy input, which is \[\eta = \frac{12.6MJ}{60.2MJ}=0.209\]to 3 significant figures.
This is the first time I've used 'heating value' for anything like this, so if the answer's wrong, I couldn't find an appropriate definition.
so, can i say that efficiancy is 20.9%?
Yes
ok, thank you, i already faned you long time ago, so cant fan again :(
Yeah I know - no worries :)
The answer makes sense - I've seen values between 10 and 50% for internal combustion engines.
then it is probably correct, i also know that cobustion engines are very unefficient
hey anwar
hi :)
want a fan?
would you be one if I say yes?
...
lol
you got it
thanks!!
anwar, can you please fan lokisan, because he helped me alot and i can only fan him once
lol you really care about others :P.. I already did anyways
so you faned me just as an exchange for me to fan the guy huh? :(
nah, would have faned you anyway
lol andy, you and your fanning job~ ^_^

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