[zerofantasym]

Hi! I've been doing this problem for an hour and I cannot figure it out.

A quantity of ice at 0 C is added to 64.3 g of water in a glass at 55 C. After the ice melted, the temperature of the water in the glass was 15 C. How much ice was added? The heat of fusion of water is 6.01 kJ/mol and the specific heat is 4.18 J/(g * C)

I keep getting 32 g, but the book says it's 27 g. Can someone tell me step by step how to do it? :S

Thanks so much!

[Borek]

Show your work.

[zerofantasym]

64.3 g H₂O x (1 mol H₂O / 18 g H₂O) x ( g * C / 4.28 J) x (6010 J / mol) x (1/40 C)

This gives me 128 g.

I figured I'm forgetting something about specific heat so I looked it up and got this


q= (4.18 J / g * C) x 64.3 g x (-40 C) = -1.08x10^4
-1.08x10^4 J x (18 g H₂O  / 1 mol H₂O) x (1/6010 J)
= 32 g


 It's been a year since I took Chemistry I. I have no idea what I'm forgetting.

[Borek]

TBH I have no idea what you are trying to do.

Start with a heat balance - heat gained (by melting and heating) is heat lost.

Heat gained must be a sum of melting and heating to the final temp, there is no sum in your calculations.

[Valdorod]

zerofantasym,

As Borek said you need to begin with the heat balance

-q_water = q_ice + q_fus

Where
q_water = mass_water * s * ΔT_water

q_ice  = mass_ice * s * ΔT_ice

q_fus = mass_ice * ΔH_fus/(Molar mass water)

put them together and solve for mass_ice and it works out to 27 g.  Just keep an eye for your signs.

Valdo