(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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

2. Relevant equations

q = mass * specific heat * (delta) temp

3. 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 thats wrong too.

Thanks for any help!!!=)

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# Homework Help: Mc(delta) t (What am I doing wrong?)

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