Heat of vaporization Physics

Question

Heat of vaporization : Q=mLv

Find what heat in calories is required to change 42 gg of 0∘C∘C ice to steam at 100∘C∘C. The heat of fusion Lf for water is 80 cal/gcal/g. The specific heat capacity of water is 1 cal/g⋅∘Ccal/g⋅∘C. The heat of vaporization for water Lv is 540 cal/gcal/g. I tried on my own and got 8400 J and it was incorrect. Not sure what to do next.

Answer 1

First you must put in enough heat to melt the ice (we can call this Q₁). Then you must put in enough heat to warm the water from 0°C to 100°C (we can call this Q₂). Then you must add enough heat to turn the liquid water into steam (we can call this Q₃).

Q_TOTAL = Q₁ + Q₂ + Q₃

Q₁ = mL_f = (42 g)(80 cal/g) = 3360 calories

Q2 = mC_pΔT = (42 g)(1 cal/(g°C))(100 - 0 °C) = 4200 calories

Q3 = mL_v = (42 g)(540 cal/g) = 22680 calories

Q_TOTAL = 3360 + 4200 + 22680 = 30,240 calories

I'll let you round to the appropriate number of sig figs.