Question

Calculate the amount of heat required to change 35.0g ice at -25.0 C to steam at 125 C (Heat of fusion = 333 J/g; heat of vaporization = 2260 J/g; Specific heats: ice = 2.09 J/g*K, water = 4.18 J/g*K, steam = 1.84 J/g*K)

Answer 1

Q = m c ∆T
Q = quantity of heat in joules (J)
m = mass of the substance acting as the environment in
grams (g)
c = specific heat capacity (4.19 for H2O) in J/(g ^oC)
∆T = change in temperature = Tfinal - Tinitial in ^oC

The "ICE" in this case is going to do 5 things:

a) as a solid, warm up from -25 to zero
b) all 35 g will melt
c) as a liquid, temperature reaches 100 Deg Cel

d) all 35 gm will convert to steam

e) as Steam it reaches temperature 125 Deg Cel

Q = 35 x 2.09 x 25 + 35 x 333 + 35 x 4.18 x 100 + 35 x 2260 + 35 x 1.84 x 125

Q = 1828.75 + 11655 + 14630 + 79100 + 8050

Q = 115263.75 Joules or 115.263 Kilo joules of heat is need  to change 35.0g ice at -25.0 C to steam at 125C