Question

How many grams of water at 50◦c must be added to 16 grams of ice at −12◦c to result in only liquid water at 0◦c?

Answer 1

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:

$q = m Cwater$ΔT$q = m (4.184\frac{J}{(g.^{0}C)} )$$(0^{0}C-50^{0}C)$

Calculating heat absorbed by 16 g of ice: Ice at $-12^{0}C$ is converted to ice at $O^{0}C$ and then ice at $O^{0} C$ to water at $0^{0}C$

$q = m Cice$ΔT + $m (Heat of fusion)$

$q = 16 g(2.11\frac{J}{g.^{0}C})(0^{0}C - (-12^{0}C))$ + $16 g (333.55\frac{J}{g})$

q = 405.12 J +5336.8 J =5741.92 J

- Heat given out by water = Heat absorbed by ice

-($m(4.184\frac{J}{g.^{0}C})(0^{0}C-50^{0}C) = 5741.92 J$

m = 27.4 g

Therefore, 27.4 g water at $50^{0}C$ must be added to 16 g of ice at $-12^{0}C$ to convert to liquid water at $0^{0}C$