Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: ILoveISO on February 07, 2010, 02:34:53 AM
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Calculate the total heat needed to convert 1.00 g of ice at -5.00 oC to water vapor at 105 oC.
Specific heat of liquid water = 4.184 J/(g oC), specific heat of solid water = 2.06 J/(g oC),
specific heat of water vapor = 1.84 J/(g oC), ΔHfusion = 6.02 kJ/mol, ΔHvap = 40.7 kJ/mol.
How come q5 = Cvapor ⋅ m ⋅ ΔT = 1.84 J/(g oC) × 1.00 g × 5.00 oC = 9.20 J
Why is it multiplied by 5 degree C? and
q3 = Cwater ⋅ m ⋅ ΔT = 4.184 J/(g oC) × 1.00 g × 100. oC = 418 J
Why is that 100 degree C?
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What is boiling point of water?
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Why is the vapor one multiplied by 5 degrees
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Think what changes must ice come through before it becomes steam at 105 deg C.
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Wouldn't it have to go thru 100 degree C before it boils? Don't get why is it multipled by only 5 degrees wouldn't it need another 100?
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Steam already has temperature of 100 deg C, so it has to be heated by 5 deg C only. Water was already heated to 100 deg C earlier.