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
`</p><p>q = m (4.184(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(J)/(g.^(0)C))(0^(0)C - (-12^(0)C))` +
`16 g (333.55(J)/(g))`

q = 405.12 J +5336.8 J =5741.92 J

- Heat given out by water = Heat absorbed by ice

-(
`m(4.184(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`