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