Calculate the heat change, in kilojoules, when 72.3 g of water in the form of steam are cooled from 110.0oC to 35.0oC. cp(ice) = 37.1 J/(mol oC)     cp(liq, H2O) = 75.3 J/(mol oC)   cp(steam) = 33.6 J/(mol oC)ΔHfus = 6.01 kJ/mol     ΔHvap = 40.67 kJ/mol Be sure you have the correct sign for qtotal and 3 sig figs.

Question
Asked Oct 14, 2019
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Calculate the heat change, in kilojoules, when 72.3 g of water in the form of steam are cooled from 110.0oC to 35.0oC.

 

cp(ice) = 37.1 J/(mol oC)     cp(liq, H2O) = 75.3 J/(mol oC)   cp(steam) = 33.6 J/(mol oC)

ΔHfus = 6.01 kJ/mol     ΔHvap = 40.67 kJ/mol

 

Be sure you have the correct sign for qtotal and 3 sig figs.

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Expert Answer

Step 1

The mass of water is 72.3g.

Molar mass of water is 18.01528g/mol.

massingram
molarmass of water 18.0152 8g/mol
72.3g
= 4.013mole
Numberof moles of water=.
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massingram molarmass of water 18.0152 8g/mol 72.3g = 4.013mole Numberof moles of water=.

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Step 2

q required to heat H2O(s) from 35oC to 0 oC, using the specific heat capacity of H2O(s).

q-37.1J/(mol°C)-(0°C-35°C)x4.013 mole=-5.211kJ
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q-37.1J/(mol°C)-(0°C-35°C)x4.013 mole=-5.211kJ

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Step 3

q required to convert 1 mol H2O(s) at 0 oC into 1 mol H2...

q-6.01kJ/mol x1 mole =6.01KJ
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q-6.01kJ/mol x1 mole =6.01KJ

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