Q = mL
We need to know the latent heat of fusion, L, for 0 degrees.
L = 336 kJ/kg, m = 2kg
where L is the latent heat for vaporization
Q = 2 * 336 = 672 kJ
For conversion between 0 and 100 degree Celsius
Q = mcθ, specific heat capacity = 4.2 kJ/kgK
Q = 2*4.2* (100 - 0) = 840 kJ
For conversion to steam at 100 degrees Celsius
Q = mL , L = latent heat of vaporization = 2256 kj/kg
Q = 2 * 2256 = 4512 kJ
Total heat = 672 + 840 + 4512 = 6024 kJ
= 6 024 000 J
But 1 calorie = 4.2 J
Therefore 6 024 000 J will be: 6 024 000/4.2 ≈ 1 434 286 Calories
≈ 1 434 kCalories
