# Chemistry Help?

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.)

Update:

(Q = ΔH vap * mole) + (Q = m *c *ΔT)

this is what im doing and im getting it wrong so idk what im doing wrong

Relevance

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

• Allan2 years agoReport

I was able to do it but I was doing it wrong the entire time. Thanks for the help

• Login to reply the answers