Calculating Gasoline Needed for 50.0hp Heat Engine Output

[ChunkymonkeyI]

Homework Statement

When gasoline(density=.7297 g/cm^3) is burned, it gives off 5.00 times 10^4 J/g(its heat of combustion). If a car's engine is 25% efficient, how much gasoline per hour must it burn in order to develop an output of 50.0hp. 1hp=746 W

Homework Equations

P=W/t
e=W/Qh

The Attempt at a Solution

I believed I was solving for time so what I did was I used P=W/t
then I rearranged it to t=W/P and plugged 5.00 times 10^4 for w and 746 times 50 for P and it didn't get me the right answer because the answer is 10.7 kg/h

 

[tiny-tim]

Hi ChunkymonkeyI!

… I used P=W/t
then I rearranged it to t=W/P and plugged 5.00 times 10^4 for w and 746 times 50 for P and it didn't get me the right answer because the answer is 10.7 kg/h

erm … what about your 25% efficiency?

 

[ChunkymonkeyI]

Hi ChunkymonkeyI! erm … what about your 25% efficiency? [/QUOTE

Do I use the 25 percent efficiency to find W and if so what do I do after that lol?

 

[tiny-tim]

Idk what to do with the 25 percent efficiency please help me!

aha!

everything in an exam question is there for a reason!

if you're totally stuck, just try either multiplying the answer by 0.25 (ie 25%), or dividing by it …

50% chance you'll get full marks! ​

in fact, the "car's engine is 25% efficient" means that the power output (to the wheels) is only 25% of the power input (from the gasoline)

 

[ChunkymonkeyI]

aha!

everything in an exam question is there for a reason!

if you're totally stuck, just try either multiplying the answer by 0.25 (ie 25%), or dividing by it …

50% chance you'll get full marks! ​

in fact, the "car's engine is 25% efficient" means that the power output (to the wheels) is only 25% of the power input (from the gasoline)

Lol but I want to learn how to do the steps can u please show me them it's bothering me that I can't solve this physics heat engine problem pleaaaaaaaaaaase

 

[tiny-tim]

the power output (to the wheels) is only 25% of the power input (from the gasoline) …

so put that into the equation

 

[ChunkymonkeyI]

the power output (to the wheels) is only 25% of the power input (from the gasoline) …

so put that into the equation

What equation

 

[tiny-tim]

(just got up :zzz: …)

What equation

this one …

… I used P=W/t
then I rearranged it to t=W/P and plugged 5.00 times 10^4 for w and 746 times 50 for P …

 

[ChunkymonkeyI]

(just got up :zzz: …)


this one …[/QUOT

But I don't know the values that I need to solve this problem can u please just show me the steps I'm not trying 2 cheat or anything I just want 2 learn how 2 do this

 

[tiny-tim]

find the energy from the gasoline,

then multiply by 0.25 to get the energy output