Answer:
Total heat ≅ 49.07 kJ
Step-by-step explanation:
Given that:
mass = 1.6 g = 0.016 kg
Initial temperature = - 16 ° C
final temperature = 112° C
specific heat for ice = 2.06 kJ/kgC
specific heat of water = 4.186 kJ/kgC
heat fusion of ice = 334 kJ/kg
specific heat for steam = 2.1 kJ/kgK
heat of vaporization of water = 2256 kJ/kg
To heat ice from -16 ° C to 0 ° C
Q₁ = 2.06 kJ/kgC × 0.016 kg × 16 ° C
Q₁ = 0.52736 kJ
To melt Ice at 0° C
Q₂= 334 kJ/kg × 0.016 kg = 5.344 kJ
To heat water from 0° C to 100° C
Q₃ = 4.186 kJ/kgC × 0.016 kg × 100° C
Q₃ = 6.6976 kJ
To vaporize water to steam at 100° C
Q₄ = 2256 kJ/kg × 0.016 kg = 36.096 kJ
To heat steam from 100C to 112° C
Q₅ = 2.1 kJ/kgC × 0.016 kg × 12 C
Q₅ = 0.4032 kJ
Total heat = Q₁ + Q₂ + Q₃ + Q₄ + Q₅
Total heat = (0.52736 + 5.344 + 6.6976 + 36.096 + 0.4032) kJ
Total heat = 49.06816 kJ
Total heat ≅ 49.07 kJ