Answer:
Heat energy needed = 3036.17 kJ
Explanation:
We have
heat of fusion of water = 334 J/g
heat of vaporization of water = 2257 J/g
specific heat of ice = 2.09 J/g·°C
specific heat of water = 4.18 J/g·°C
specific heat of steam = 2.09 J/g·°C
Here wee need to convert 1 kg ice from -13°C to vapor at 100°C
First the ice changes to -13°C from 0°C , then it changes to water, then its temperature increases from 0°C to 100°C, then it changes to steam.
Mass of water = 1000 g
Heat energy required to change ice temperature from -13°C to 0°C
H₁ = mcΔT = 1000 x 2.09 x 13 = 27.17 kJ
Heat energy required to change ice from 0°C to water at 0°C
H₂ = mL = 1000 x 334 = 334 kJ
Heat energy required to change water temperature from 0°C to 100°C
H₃ = mcΔT = 1000 x 4.18 x 100 = 418 kJ
Heat energy required to change water from 100°C to steam at 100°C
H₄ = mL = 1000 x 2257 = 2257 kJ
Total heat energy required
H = H₁ + H₂ + H₃ + H₄ = 27.17 + 334 + 418 +2257 = 3036.17 kJ
Heat energy needed = 3036.17 kJ
, to check it.