The energy required to melt one gram of ice at 0 degrees is given by:
[tex]Q=m L_f[/tex]
where m is the mass of the ice and [tex]L_f = 334 J/g[/tex] is the latent heat of fusion of ice. Substituting, we find
[tex]Q=(1g)(334 J/g)=334 J[/tex]
The energy required to boil one gram of water at 100 degrees is given by:
[tex]Q=m L_e[/tex]
where m is the mass of the water and [tex]L_e = 2265 J/g[/tex] is the latent heat of evaporation of the water. Substituting, we find
[tex]Q=(1 g)(2265 J/g)=2265 J[/tex]
Therefore, it is required more energy to boil one gram of water at 100 degrees than to melt one gram of ice at 0 degrees.
