An ice cube of

6.50g

at

-3.20\\\\deg C

is added to a cup with

215.0g

water at

25.10\\\\deg C

. Calculate the temperature of the water in

\\\\deg C

once equilibrium has been achieved. Assume the absence of any heat exchange of the contents of the cup with the environment.\

C_(ice )=2.09Jg^(-1)K^(-1);C_(water )=4.18Jg^(-1)K^(-1);\\\\Delta _(fusion )H=6.01kJmol^(-1);M_(water )=\ 18.02gmol^(-1)

\ Enter your numeric answer as a decimal number (e.g. 1.23). Do not use scientific notation!\ Your Answer:\ Answer\ units

An ice cube of 6.50g at 3.20C is added to a cup with 215.0g water at 25.10C. Calculate the temperature of the water in C once equilibrium has been achieved. Assume the absence of any heat exchange of the contents of the cup with the environment. Cice=2.09Jg1K1;Cwater=4.18Jg1K1;fusionH=6.01kJmol1;Mwater=18.02gmol1 Enter your numeric answer as a decimal number (e.g. 1.23). Do not use scientific notation! Your Answer: Answer units