Answer to Question #204472 in Mechanics | Relativity for Master j

Question

A 500 g bar of ice is initially at -25 °C. Calculate the amount of energy required to vapourize 25% of the substance at 100 °C.

Answer 1

The specific heat capacities are following:

ice: 2.093 J/g^oC

water: 4.184 J/g^oC

The heat of fusion of H₂O (L_ice) = 333.55 J/g

The heat of vaporization of H₂O (L_steam) = 2256 J/g

Heat required to heat the ice from -25^oC to the melting (0^oC):

$Q_1=c_{ice}m\Delta{T}=2.093J/g\cdot{\degree{C}}\times500g\times(0\degree{C}-(-25\degree{C}))=26163J$

Heat required to melt the ice at 0^oC:

$Q_2=L_{ice}m=333.55J/g\times500g=166775J$

Heat required to heat the water from 0^oC to the boiling (100^oC):

$Q_3=c_{water}m\Delta{T}=4.184J/g\cdot{\degree{C}}\times500g\times(100\degree{C}-0\degree{C})=209200J$

Heat required to evaporate 25% of water (which is $500g\times0.25=125g$ ) at 100^oC:

$Q_4=L_{steam}m=2256J/g\times125g=282000J$

The total heat is the sum of the heats required for each step:

$Q=Q_1+Q_2+Q_3+Q_4=26163J+166775J+209200J+282000J=684138J=684kJ$

Answer: 684 kJ