The heat of fusion for water at its normal freezing or meltingpoint is 333 kJ/kg
ID: 1754145 • Letter: T
Question
The heat of fusion for water at its normal freezing or meltingpoint is 333 kJ/kg and the heat of vaporization is 2256 kJ/kg the specific heat for water is 4180 What mass of steam at 100 degrees Celcuis must be mixed with150 gram ice at its melting point, in a thermally insulatedcontainer, to produce liquid water at 50 degreesCelcius???????? The heat of fusion for water at its normal freezing or meltingpoint is 333 kJ/kg and the heat of vaporization is 2256 kJ/kg the specific heat for water is 4180 What mass of steam at 100 degrees Celcuis must be mixed with150 gram ice at its melting point, in a thermally insulatedcontainer, to produce liquid water at 50 degreesCelcius????????Explanation / Answer
The heat of fusion for water at its normal freezing or meltingpoint is 333 kJ/kg and the heatof vaporization is 2256 kJ/kg. the specific heat for water is 4180.What mass of steam at 100
degrees Celsius must be mixed with 150 gram ice at its meltingpoint, in a thermally insulated
container, to produce liquid water at 50 degrees Celsius?
solution: Let x kg of steam be needed.
heat lost by steam = 2256x + 4180x(50) = heat gained by ice =.15(333) + .15(4180)(50)
2256000x + 4180x(50) = .15(333000) + .15(4180)(50)
now you can solve for x.
hope this helps!
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