The amount of energy released by burning a fuel source, measured in energy per mass, is called the fuel value. If all the energy obtained from burning 1.28 pounds of methane (fuel value is 11.97 kcal/g) is used to heat 127.0 kg of water at an initial temperature of 21.5 °C, what is the final temperature? help??

Answer 1

1.28 lbs x 453.59 g/lb = ?grams CH4.
?grams CH4 x (11.97 cal/g) = ? cal heat produced by the l.28 lbs CH4.

?cal = mass H2O x specific heat H2O x (Tfinal-Tintial)
Substitute 127,000 for mass H2O
Solve for Tf
Substitute 21.5 for Ti
Substitute 1 cal/g for specific heat H2O.

Answer 2

I keep getting the wrong answer :/
I did:
1.28lb x 453.59g/lb= 580.5952g CH4
580.5952g Ch4 x 11.97 cal/g= 6949.724544 cal
6964.724544 cal= 127,000 x 1 cal/g x (Tf-21.5)
Final Temp= 21.6*C
but this answer was wrong

Answer 3

nvm! i got it! I forgot to convert 11.97 kcal/g to cal/g

Answer 4

1.28lb x 453.59g/lb= 580.5952g CH4
580.5952g Ch4 x 11.97 cal/g= 6949.724544 cal
6964.724544 cal= 127,000 x 1 cal/g x (Tf-21.5)
You are ok to here. You didn't show your next step but that is the error. Here is your last line.
6964.72 = 127 x 1 x (Tf-21.5)
Then
6964.72 = 127Tf - (127*1*21.5)
6964.72 = 127Tf - 2730.50
6964.72 + 2730.50 = 127Tf
I'll let you take it from here.
Final Temp= 21.6*C

Answer 5

I forgot and didn't turn off the bold and it printed hat last line in bold. Ignore "Final Temp = 21.6"