An ice cube has mass 9.0g and is added to a cup of coffee. the coffee's intial temp. is 90.0 deg. celsius and the cup contains 120.0 g of liquid. assume the specific heat capacity of coffee is the same as that of water. the heat of fusion of ice (associated with ice melting) is 6.0kj/mol. find the temp. of coffee after ice melts.

Are we using deg C or deg F? but i'll assume its deg Cto get rid of the mol, just find the molecular mass of icewhich is H2O, 18g/molso therefore, specific latent heat of fusion of ice = 6.0kJ/18g                                                  = 333.3J/glatent heat of fusion of ice = (6.0kJ/18)*9                                      = 3.0kJok i am not too sure what value you are using for specific heat capacity of water, but lets just use 4.20J/gtherefore3.0kJ = 120*c*(change in temperature)         = 504* change in temperaturechange in temperature = 3000/504                                  = 5.952 KHence, final temperature = 90.0 - 5.952                                     = 84.0 deg C

Hey how did you manage to type such a big lab report