Asked by Dakota
The heat of fusion of water is 335 J/g, the heat of vaporization of water is 2.26 kJ/g, and the specific heat of water is 4.184 J/deg/g. How many grams of ice at 0 ° could be converted to steam at 100 °C by 9,946 Joules of heat?
Answers
Answered by
DrBob222
Let Y = grams ice.
335*Y + (Y*4.184*100) + 2226*Y = 9946.
Solve for Y in grams.
335*Y + (Y*4.184*100) + 2226*Y = 9946.
Solve for Y in grams.
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