Do it for 1 kilogram first, then multiply by 0.003
1. melt the ice
heat in = 336 kJ
2. raise temp of water from 0 to 100
heat in = 100 (4.2 kJ/degK) = 420 kJ
3. boil it
heat in = 2260 kJ
add
3016 kJoules for 1 kg
multiply by 0.003
9.05 kJoules = 9050 Joules
How much heat energy is needed to change 3g of
ice at 0C to steam at 100C(Specific latent heat of
fusion of ice=336kj/kg, specific latent of
vapourization of water=2260kj/kg, specific heat
capacity of water=4.2kj/kgk)?
1 answer