An ice cube of

6.00g

at

-4.50\\\\deg C

is added to a cup with

285.0g

water at

27.20\\\\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!

An ice cube of 6.00g at 4.50C is added to a cup with 285.0g water at 27.20C. 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