Posted by **chirayu** on Wednesday, March 25, 2009 at 7:15am.

What mass (in grams) of steam at 100°C must be mixed with 340 g of ice at its melting point, in a thermally insulated container, to produce liquid water at 12.0°C? The specific heat of water is 4186 J/kg·K. The latent heat of fusion is 333 kJ/kg, and the latent heat of vaporization is 2256 kJ/kg

- thermal physics -
**drwls**, Wednesday, March 25, 2009 at 9:02am
Let the unknown mass of steam required by X grams. The amount of heat that is loses changing to 12 C liquid water equals the heat gained by the 340g of ice as it melts to form water at the same temperature (12 C). If we use units of J/g and J/g K,

X*[2256 + 4.186(100 - 12)]

= 340 [333 + (12 - 0)4.186)]

2624 X = 130,299

Solve for X

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