Calculating Final Temperature of Mixed Ice and Water Sample | No Heat Loss

[meganw]

Homework Statement

A 19.2 g sample of ice at -10.0°C is mixed with 100.0 g of water at 74.9°C. Calculate the final temperature of the mixture assuming no heat loss to the surroundings. The heat capacities of H2O(s) and H2O(l) are 2.08 and 4.18 J g-1 °C-1, respectively, and the enthalpy of fusion for ice is 6.02 kJ/mol.
__________ degrees Celsius

Homework Equations

q = mass * specific heat * (delta) temp

The Attempt at a Solution

I thought I knew how to do it but keep getting it wrong. =(

19.2 g ice at -10.0 deg going to 0 deg:
19.2g x 2.08 J/g deg x 10 deg = 399.96 J
Melting 19.0 g ice at 0 deg requires-
19.2g x (1/18.02 g/mol) x 6.02kJ/mol x 1000 J/kJ x = 6350.77 J
399.96 + 6350.77 = 6750 J
That energy will come from cooling 100.0 g H20:
6750 = (100)(4.18)(T-74.9)
T(final) = 91.04

thats wrong on webassign though...

I also tried doing 74.9 - T to see if that works and I get 58.8 but that's wrong too.

Thanks for any help! =)

 

[Borek]

Once the ice gets melted you have a mixture of 19.2g water at 0 deg C and 100g water at some other temperature. Neither of these temperatures is final.

Also note that 91.04 is obviously wrong - you can't take a water at 74.9 deg C, add ice to it and end with water that is hotter than it was before.

 

[meganw]

Hmmm...then how would I relate the temps if they are not final? What else would I need to do?? :-/

 

[Borek]

You know that cold water has to get warmer and the cold one has to get colder. Heat gained = heat lost, they have both identical final temperature.