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.

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Expert Solution

Answer - Given, mass of water = 455 g , ti = 80.0^oC, tf = 100.0^oC

Specific heat of water = 4.184 J/g^oC , ∆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/g^oC * (100.0-80.0)^oC

= 38074.4 J

Now heat from the 100.0 to 100.0^oC

q2 = m*∆Hvap

= 18518.5 J

So, total heat absorbed = q1 + q2

= 38074.4 + 18518.5

= 56593 J

= 56.59 kJ

queen_honey_blossom answered 1 year ago

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