How much steam must be passed into a mixture of ice and water in order to melt 10g of ice.?

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answers:
The original Peter G say: about 1.48 grams
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david say: 10 X 80 = 800 cal to melt the ice
m = mass steam
540m + 100m = 640m <<< cal released by the steam
heat loss = heat gain
640m = 800
m = 800/640 = 1.25g steam
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Dr W say: not solvable without knowing properties of the steam (pressure, temperature, etc)
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Roger the Mole say: (10 g ice) x (333.6 J/g) = 3336 J required

Supposing the steam to be initially at 100°C:
2257 J/g from condensing the steam
(4.184 J/g·°C) x (100 - 0)°C = 418.4 J/g from cooling the condensed steam to 0°C
2257 J/g + 418.4 J/g = 2675.4 J/g total

(3336 J) / (2675.4 J/g) = 1.2 g steam
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