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How much heat is required to convert 4.88 g of ice at − 13.0 ∘C to water at 28.0 ∘C ?
(The heat capacity of ice is 2.09 J/g∘C, ΔHvap(H2O)=40.7kJ/mol, ΔHfus(H2O)=6.02kJ/mol)
1 Answer
- billrussell42Lv 73 years agoFavorite Answer
energy to warm ice to 0º
E1 = 2.06 J/gC x 4.88 g x 13C
energy to melt ice
E2 = 334 J/g x 4.88 g
energy to warm water to 28
E3 = 4.186 J/gC x 4.88 g x 28C
add the three to get the answer.
specific heat of water is 4.186 kJ/kgC = 4.186 J/gC
specific heat of ice is 2.06 kJ/kgC
specific heat of steam is 2.1 kJ/kgK
heat of fusion of ice is 334 kJ/kg
