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My name is Ann [436]
3 years ago
15

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?

Chemistry
1 answer:
melisa1 [442]3 years ago
4 0

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 CwaterΔTq = 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 CiceΔT + 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

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