Question
In: Chemistry
how many kj of heat are absorbed when 455g of water at 80.0C are heated to...
how many kj of heat are absorbed when 455g of water at 80.0C are heated to 100.0C and then completely evaporated at this tempertature? The Specific heat of water 4.184j/C, g and the molar heat of evaporation for water is 40.7kJ/mol.
Solutions
Expert Solution
Answer - Given, mass of water = 455 g , ti = 80.0oC, tf = 100.0oC
Specific heat of water = 4.184 J/goC , ∆Hvap = 40.7 kJ/mol
Moles of water = 455 g / 18.016 g.mol-1
= 25.3 moles
We know the formula for the heat
q1 = m*C*∆t
= 455 g * 4.184 J/goC * (100.0-80.0)oC
= 38074.4 J
Now heat from the 100.0 to 100.0oC
q2 = m*∆Hvap
= 18518.5 J
So, total heat absorbed = q1 + q2
= 38074.4 + 18518.5
= 56593 J
= 56.59 kJ
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