the heat of fusion water is 335J/g,the heat of vaporization of water is 2.26kJ/g, the specific heat of water is 4.184J/deg/g. how many grams of ice at 0 degrees could be converted to steam at 100 degrees C by 9,574J

Answer 1

Let x = mass ice.
335*x = heat needed to melt ice.
4.184*100*x = heat needed to raise T of that ice from zero C to 100 C.
2260x = heat needed to vaporize the ice at 100 to steam at 100.

335x + (4.184*100*x) + 2260x = 9574 J. Solve for x.