[Dolphinsiu]

I get stuck when doing Q.1

Theoretically, how high could a gallon of gasoline lift an automobile weighing 2800 lb against the force of gravity, if it is assumed that the cylinder temperature is 2200 K and the exit temperature 1200 K ? (Density of gasoline = 0.80 g/cm3 ; 1 lb = 453.6 g; 1 ft = 30.48 cm; 1L = 0.2642 gal: heat of combustion of gasoline = 46.9 kJ/g)
Ans. 17000 ft

I have attempted as follows:

Heat loss by gasoline = potential energy gained by automobile

Heat loss by gasoline:
(46.9)(1000)[1 gal (1L /0.2642 gal) (1000 cm3/1L) (0.8 g/cm3)
= 9912784 J

Potential energy gained by automobile:
(2800 lb) (453.6 g/1 lb) (1 kg/1000g) (9.80655) h

but I find that h = 37 ft but not 17000 ft, why?

Do the question itself have problem? because I don't know why it emphasis on temperature changes from 2200 K to 1200 K!

[enahs]

Well to get you started,your first calculation is wrong.

With a density of 0.8g/mL, there is 3,028.4 g in a gallon.
46.9 * 3028.4 = 142028.7 kJ of energy released.

If we did it your way, that would give you 37 thousands+ ft.


But that assumes no energy lost in the form of heat (q).
This is not true, as you are told so in the question.

You have to use the fact that
U=q+w
You know U (total energy of the system). You have to figure out q, and subtract that from U to get the energy in the form of work done, and then use that to calculate the height it can lift it.

Try that and see if it gets the correct answer.

[Borek]

http://en.wikipedia.org/wiki/Heat_engine#Efficiency

[Dolphinsiu]

Yes
My calculation is wrong!

But how do I know the Cp value of gasoline? as I truly know that

qv = Cv delta T as the volume of container is constant.

Can I treat Cp value of gasoline as that of methanol (found from the last pages of book)?

[enahs]

For an ideal gas, C_V,m= 3/2 * R and C_P,m = 5/2 * R


However, I am now having some problems after reading the question fully. The method I suggested was not for a reversible heat engine but just an expansion due to heat of conbustion, but the answer is based off of a reversible heat engine. That means you work it as Borek suggested; which is much easier anyway. The simplified efficiency equation takes care of q in the type of condition that you apparently have, despite it was not stated.


So once you have the correct first number calculated, and the efficiency of the engine, it is then working it just like you were.

[Dolphinsiu]

Thank you!