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Calculate the amount of heat in kilojoules required when 31.0 g of water at 52.5°C is...

Calculate the amount of heat in kilojoules required when 31.0 g of water at 52.5°C is converted to steam at 238.2°C.

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Answer #1

Ti = 52.5 oC

Tf = 238.2 oC

here

Cl = 4.184 J/goC

Heat required to convert liquid from 52.5 oC to 100.0 oC

Q1 = m*Cl*(Tf-Ti)

= 31 g * 4.184 J/goC *(100-52.5) oC

= 6160.94 J

Lv = 2260.0 J/g

Heat required to convert liquid to gas at 100.0 oC

Q2 = m*Lv

= 31.0g *2260.0 J/g

= 70060 J

Cg = 2.01 J/goC

Heat required to convert vapour from 100.0 oC to 238.2 oC

Q3 = m*Cg*(Tf-Ti)

= 31 g * 2.01 J/goC *(238.2-100) oC

= 8611.242 J

Total heat required = Q1 + Q2 + Q3

= 6160.94 J + 70060 J + 8611.242 J

= 84832 J

= 84.8 KJ

Answer: 84.8 KJ

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