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.

How do we get rid of the /mol under the kJ?
Also, may I get a hint please?

Question

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.

How do we get rid of the /mol under the kJ?
Also, may I get a hint please?

anonymous · Answer

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