When water at 50 C is added to ice at -12 C, heat is transferred from hot water to ice.
- Heat given out by water = Heat absorbed by ice
Calculating the heat released by hot water:
[tex] q = m Cwater [/tex]ΔT[tex]
q = m (4.184\frac{J}{(g.^{0}C)} ) [/tex][tex] (0^{0}C-50^{0}C) [/tex]
Calculating heat absorbed by 16 g of ice: Ice at [tex] -12^{0}C [/tex] is converted to ice at [tex] O^{0}C [/tex] and then ice at [tex] O^{0} C [/tex] to water at [tex] 0^{0}C [/tex]
[tex] q = m Cice [/tex]ΔT + [tex] m (Heat of fusion) [/tex]
[tex] q = 16 g(2.11\frac{J}{g.^{0}C})(0^{0}C - (-12^{0}C)) [/tex] + [tex] 16 g (333.55\frac{J}{g}) [/tex]
q = 405.12 J +5336.8 J =5741.92 J
- Heat given out by water = Heat absorbed by ice
-([tex] m(4.184\frac{J}{g.^{0}C})(0^{0}C-50^{0}C) = 5741.92 J[/tex]
m = 27.4 g
Therefore, 27.4 g water at [tex] 50^{0}C [/tex] must be added to 16 g of ice at [tex] -12^{0}C [/tex] to convert to liquid water at [tex] 0^{0}C [/tex]
