2011-04-12, 09:21 PM
Okay I looked up some constants. I hope they're right.
Q = mst
T is the change in temperature. 113 - 30 = 83 Celsius.
Specific heat of water is 4.186.
Mass is 377 grams.
Q = (83) (377) (4.186)
Q = 130984 Joules = 130.98 kilo Joules
Now we need to find a constant that methane burns at. I'm not too sure about this one. I looked at the back of my book and got the heat of FORMATION to be -74.8. But I also googled it and got that it could be 55. Both in kilojoules.
So we do 130.98/74.8 (or 130.98/55) to get the number of moles of methane needed.
I get 1.75 moles (or 2.38 moles).
Then to get grams. Carbon weighs 12 grams, 4 x Hydrogen weighs 1 gram. 16 grams total.
16 Molecular mass x 1.75 moles = 28 grams. (Or 2.38 x 16 = 38.08 grams)
So 28 or 38. Maybe. I'm not sure, this question gives little information.
Q = mst
T is the change in temperature. 113 - 30 = 83 Celsius.
Specific heat of water is 4.186.
Mass is 377 grams.
Q = (83) (377) (4.186)
Q = 130984 Joules = 130.98 kilo Joules
Now we need to find a constant that methane burns at. I'm not too sure about this one. I looked at the back of my book and got the heat of FORMATION to be -74.8. But I also googled it and got that it could be 55. Both in kilojoules.
So we do 130.98/74.8 (or 130.98/55) to get the number of moles of methane needed.
I get 1.75 moles (or 2.38 moles).
Then to get grams. Carbon weighs 12 grams, 4 x Hydrogen weighs 1 gram. 16 grams total.
16 Molecular mass x 1.75 moles = 28 grams. (Or 2.38 x 16 = 38.08 grams)
So 28 or 38. Maybe. I'm not sure, this question gives little information.