[URGENT!!] How much energy (heat) is required to convert 25.5 g of H2O(l) at 35.0°C to H2O(g) at 115.0°C?
specific heat of ice: 2.09 J/g·°C
Δ Hfus = 6.02 kJ/mol
specific heat of water: 4.18 J/g·°C
Δ Hvap = 40.7 kJ/mol
specific heat of steam: 1.84 J/g·°C
More Questions From This User See All
Answers & Comments
(4.18 J/g·°C) x (25.5 g) x (100 - 35.0)°C = 6928.35 J to warm the water to its boiling point
(40.7 kJ/mol) x (25.5 g H2O / (18.01532 g H2O/mol)) = 57.60930 kJ = 57609.30 J to vaporize the liquid
(1.84 J/g·°C) x (25.5 g) x (115.0 - 100)°C = 703.8 J to heat the steam to 115.0°C
6928.35 J + 57609.30 J + 703.8 J = 65241 J = 65.2 kJ total