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Answer to Question #180689 in Physical Chemistry for Mike Chow
Question #180689
How much heat energy is required to convert 200.0 g of ice at −20 oC to all steam at 110 oC at 1 atm?
Expert's answer
q = q1 + q2
q1= Mc∆T
∆T = 110--20 = 130°C
= 200 × 4.184J/C/g × 130
= 108784J
q2 = M∆Hvap
= 200 × 2260J/g
= 452000
= 108784 + 452000
= 560785/100
= 560.784KJ
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