Convert 1.29 lbs C4H10 to grams = ?
Then ? g x 10.85 kcal/g = q kcal or 1000q cal.
1000q = mass H2O x specific heat H2O x (Tf-Tinitial)
Solve for Tf.
I estimate the answer is about 71 C.
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 1.29 pounds of butane (fuel value is 10.85 kcal/g) is used to heat 122.5 kg of water at an initial temperature of 18.8 °C, what is the final temperature?
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