can some help me with this
How much heat is needed to convert 863 g of ice at -30°C to steam at 123°C? (The specific heats of ice, steam, and liquid are 2.03 J/g·°C, 1.99 J/g·°C, and 4.18 J/g·°C, respectively. The heat of fusion is 6.02 kJ/mol and heat of vaporization is 40.79 kJ/mol.)
(Q = ΔH vap * mole) + (Q = m *c *ΔT)
this is what im doing and im getting it wrong so idk what im doing wrong
- Roger the MoleLv 72 years agoFavorite Answer
Are you paying attention to when to use grams and when to use moles?
(863 g H2O) / (18.01532 g H2O/mol) = 47.9037 mol H2O
(2.03 J/g·°C) x (863 g) x (0 - (-30))°C) = 52556.7 J = 52.557 kJ to warm the ice to its melting point
(6.02 kJ/mol) x (47.9037 mol) = 288.380 kJ to melt the ice
(4.18 J/g·°C) x (863 g) x (100 - 0)°C = 360734 J = 360.734 kJ to warm the melted ice to its boiling point
(40.79 kJ/mol) x (47.9037 mol) = 1953992 J = 1953.992 kJ to vaporize the water
(1.99 J/g·°C) x (863 g) x (123 - 100)°C = 39500 J = 39.500 kJ to heat the steam to 123°C.
52.557 kJ + 288.380 kJ + 360.734 kJ + 1953.992 kJ + 39.500 kJ = 2695 kJ total