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What mass of steam at 100 degrees C must be added to 2.15kg of ice at...

What mass of steam at 100 degrees C must be added to 2.15kg of ice at 0 degrees C to yield liquid water at 30 degrees C?

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Answer #1

  Heat required converting water at 0° C to ice at 0° C

= 2.15*333550 = 717132.5 J.
Heat required raising its temperature of water from 0° C to 30 ° C
= 2.15*4186*30 =269997

Total heat required = 987129.5J
--------------------------------------...
Heat to be removed from steam at 100° C to convert into water at 100° C
= 2260000* M.
Heat to be removed to cool water at 100° to water at 16 ° C

= M*4186* [100 -30] =293020 M
Total heat to be removed = 2525555 M J
--------------------------------------...
2525555 M = 987129.5

M = 0.138kg of steam

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