calculate the total amount of heat required to convert hundred gram of ice at -5 degree centigrade into "steam" at 100°C.
Answers
Heat the ice from -30 to 0
30 * 2090 J
melt the ice
333000 J
heat the resulting water 100 degrees
100 * 4180 Joules
boil the water
2,260,000 Joules
heat the steam
(120-100)(2,010) Joules
Add them up and divide by 1000 to get a gram instead of a kilogram
Heat required = ΔH(fusion) + ΔH(vaporisation) + heat required to rise temperature of ice to 0°c + heat required to rise temperature of water to 100°c
(mass = 100g=0.1kg)
( Lf = latent heat of fusion = 3.35x 10^5 J/kg
Lv= latent heat of vaporisation = 2.268 x 10^6 J/kg
specific heat of water = Cw = 4.2 x 10^3 J/kg
specific heat of ice = Ci = 2.04 x 10^3 )
heat required = mCiΔT + mCwΔT + mLf + mLv
= 0.1 ( 5Ci + 100Cw + Lf + Lv)
= 30.332 x 10^4 = 3.0332 x 10^5 J
( the calculations are rough , pls recheck )