posted by Anonymous on .
the molar heat of combustion of propane is -2.2*10^3 kJ. how many grams of propane would have to combusted to raise 1.0kg of water from 22.0 C to 100.0 C?
not sure where to start please someone help
How much heat do you need to heat the water?
q = mass H2O x specific heat H2O x delta T = 1000g x 4.184 J/g*C x (100-22) =? J.
2200 J/mol x ?mol = qJ (from above)
Solve for ?mol propane and convert to grams.
So I did what you told me and the answer I got is 6541 grams but the book key answer is 6.5 grams
You're right. I didn't change 2.23E3 kJ to J although I thought I did when I typed in 2200.
q = 1000 x 4.184 x (100-22).
2200000 x ?mol = q
?mol x 44 = 6.527 grams which rounds to 6.5 to two s.f.
Thank You so much for your help!!for this question I was just wondering if you could see if i did it right"the specific heat of water is 4.184 j/g °c. How many grams of water at 85 C would have to be added to raise 1.00 kg of water from 25.0 C to 50.0 C?" for my answer I got 714 grams of water but I am not sure if that is correct.
Didn't Bob Pursley work that problem for you earlier? I would take his answers to the bank.