[mookxi]

I'm not really sure how to go about doing this question. Any clues? ;-)

A 300mL volume of LP gas can be burnt to heat 400mL of water from 18.0 degrees celcius to 34.7 degrees celcius. What is the heat of combustion of the LP gas in kJ / L?



thanks ;-)

[Borek]

Simple heat balance.

[sdekivit]

A 300mL volume of LP gas can be burnt to heat 400mL of water from 18.0 degrees celcius to 34.7 degrees celcius. What is the heat of combustion of the LP gas in kJ / L?

300 mL: Q = (c * m * delta T)_water

--> now you know heat of combustion of 300 mL LP so you need to convert it to Q / L (assuming no heat is lost to the environment)

[anarchron]

Check out this answer:

as attachment

[mookxi]

I first and second answer didn't make sense to me, but the third did! thanks.

[Borek]

I first and second answer didn't make sense to me, but the third did! thanks.

So you have a problem, because anarchron answer is wrong.

[sdekivit]

So you have a problem, because anarchron answer is wrong.

if the mass of water is 400 g instead of the 300 g used in his calculation, the answer will be correct

[Borek]

But only then

Besides, funny thing is that anarchron answer does exactly what we have both suggested, just refraining from giving out the detailed solution.

[sdekivit]

But only then

Besides, funny thing is that anarchron answer does exactly what we have both suggested, just refraining from giving out the detailed solution.

true 

[mookxi]

But only then

Besides, funny thing is that anarchron answer does exactly what we have both suggested, just refraining from giving out the detailed solution.

Oh really? I'm sorry! I suppose the fact that we're both doing the same course study and method helps. I didn't quite understand what "Q", "c*" means.

And oh, I didn't even notice that Anachron's answer was wrong, Thanks for picking it up. I used 400g anyway ;-)