How many grams of gasoline would you need to burn if you started with a 275 gram piece of ice that started at -25 OC and by the end of the experiment had 100 grams of steam and the rest liquid water?

Answer 1

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This is a many step problem. How much have you done and exactly what do you not understand.

Answer 2

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(2.06)(275)(0-(-25))=14162.5

333J/100x14162.5 = 47161.125

47161.125+14162.5= 61323.625

61323.625(1gas/48,000)= answer: 1.277g of steam

Thank you to the way for helping chemistry isn't my best subject so I apreciate the time y'all put in to help tutor

Answer 3

CpxmassxTf-Ti
(2.06)(275)(0-(-25))=14162.5 This step is perfect. Good work!

333J/100x14162.5 = 47161.125 I don't understand this step. You should be melting all 275 g of the ice so it should be 333 J/g x 275 g = ? J for the melting step.

Now you want to raise the temperature of the 275 g water from zero C to 100 C which is 275g x 4.184 J/g x (100-0) = ?

Then you want to vaporize 100 g of the 275 g H2O you have to steam which will leave you with 175 g of liquid water.
That is 100 g H2O x heat vaporization which I think is about 2260 J/g (but check me out on that for sometimes I don't remember all of these constants) = ? J.
All all of those values together to find total q needed.

Then look up the heat of combustion of gasoline and we can talk about that step when you get there.

47161.125+14162.5= 61323.625

61323.625(1gas/48,000)= answer: 1.277g of steam