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Calculate the amount of energy in kilojoules needed to change 153g of water ice at −10 ∘C to steam at 125 ∘C.?
Calculate the amount of energy in kilojoules needed to change 153g of water ice at −10 ∘C to steam at 125 ∘C. The following constants may be useful:
Cm (ice)=36.57 J/(mol⋅∘C)
Cm (water)=75.40 J/(mol⋅∘C)
Cm (steam)=36.04 J/(mol⋅∘C)
ΔHfus=+6.01 kJ/mol
ΔHvap=+40.67 kJ/mol
My answer is way wrong if you can help me that would be great! Thanks
I took 0 C to -10
10*36.57+6.01+75.40*100+40.67+36.04*25
1 Answer
- Roger the MoleLv 76 years ago
Yes, your answer is wrong. You have ignored the units completely.
Since all the constants are given in moles, convert the mass of water to moles:
(153 g H2O) / (18.01532 g H2O/mol) = 8.4928 mol H2O
(36.57 J/(mol⋅∘C)) x (8.4928 mol) x (0 - (-10)) ∘C = 3105.82 J to warm the ice to its melting point
(6.01 kJ/mol) x (8.4928 mol) = 51.04173 kJ = 51041.73 J to melt the ice
(75.40 J/(mol⋅∘C)) x (8.4928 mol) x (100 - 0) ∘C = 64035.71 J to warm the water to its boiling point
(40.67 kJ/mol) x (8.4928 mol) = 345.40218 kJ = 345402.18 J to vaporize the water to steam
(36.04 J/(mol⋅∘C)) x (8.4928 mol) x (125 - 100) ∘C = 7652.01 J to warm the steam to 125∘C
3105.82 J + 51041.73 J + 64035.71 J + 345402.18 J + 7652.01 J = 471237.45 J = 471 kJ total
