Answer:
46.85kJ
Explanation:
The specific latent heat of vaporization of water is =2.3 x 10⁶ J kg-1.
The specific heat capacity of water = 4.2×10³J/kgK
The latent heat of fusion of water= 3.36×10⁵ J/kg
The specific heat capacity of ice=2.108×10³ J/kgK
The heat of lost due to cooling of ice is calculated as follows by adding the heat lost during condensation of water vapor to the heat lost during cooling the water to freezing point plus the heat lost during freezing plus the heat lost during cooling the ice to -32.0°C
=MLv+MCΔT(water)+MLf+MCΔT(ice) where Lv=latent heat of vaporization of water , and Lf is the latent heat of fusion ice.
=(0.015kg×2.3 x 10⁶ J/kg)+(0.015kg×4.2×10³J/kgK×100K)+(0.015kg×3.36×10⁵ J/kg)+(0.015kg× 2.108×10³ J/kgK)×32K
=34500J+6300J+5040J+1011.84J
=46851.84J
=46.85kJ
