Calculate the amount of energy in kilojoules needed to change 153 g of water ice at...

Calculate the amount of energy in kilojoules needed to change 153 g of water ice at –10 ∘C to steam at 125 ∘C. The following constants may be useful: Cm (ice)=36.57 J mol−1 ∘C−1 Cm (water)=75.40 J mol−1 ∘C−1 Cm (steam)=36.04 J mol−1 ∘C−1 ΔfusH=+6.01 kJ mol−1 ΔvapH=+40.67 kJ mol−1

Expert Solution

Ti = -10.0 oC
Tf = 125.0 oC
here
Cs = 36.57 J/mol.oC

Lets convert mass to mol
Molar mass of H2O = 18.016 g/mol
number of mol
n= mass/molar mass
= 153.0/18.016
= 8.4925 mol

Heat required to convert solid from -10.0 oC to 0.0 oC
Q1 = n*Cs*(Tf-Ti)
= 8.4925 mol * 36.57s J/mol.oC *(0--10) oC
= 3105.6894 J

Hfus = 6.01KJ/mol =
6010J/mol

Heat required to convert solid to liquid at 0.0 oC
Q2 = n*Hfus
= 8.4925 mol *6010 J/mol
= 51039.6314 J

Cl = 75.4 J/mol.oC

Heat required to convert liquid from 0.0 oC to 100.0 oC
Q3 = n*Cl*(Tf-Ti)
= 8.4925 mol * 75.4 J/mol.oC *(100-0) oC
= 64033.0817 J

Hvap = 40.67KJ/mol =
40670J/mol

Heat required to convert liquid to gas at 100.0 oC
Q4 = n*Hvap
= 8.4925 mol *40670 J/mol
= 345387.9885 J

Cg = 36.04 J/mol.oC

Heat required to convert vapour from 100.0 oC to 125.0 oC
Q5 = n*Cg*(Tf-Ti)
= 8.4925 mol * 36.04 J/mol.oC *(125-100) oC
= 7651.6985 J

Total heat required = Q1 + Q2 + Q3 + Q4 + Q5
= 3105.6894 J + 51039.6314 J + 64033.0817 J + 345387.9885 J + 7651.6985 J
= 471218 J
= 471.2 KJ
Answer: 471.2 KJ

queen_honey_blossom answered 1 year ago

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