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?


1
Expert's answer
2021-04-13T08:04:21-0400

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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