Thermal equilibrium with water and ice

[louisfrancois]

A 3.8 kg block of copper at a temperature of 72C is dropped into a bucket containing a mixture of ice and water whose total mass is 1.2 kg. when thermal equilibrium is reached, the temperature of the water is 8C. How much ice was in the bucket before the copper block was placed in it? (Neglect heat capacity of the bucket).

Heat gained=heat lost
Mice(Lf)+(Mwater)(Cwater)(Teq-Ti)=(Mcopper)(Ccopper)(Ti-Teq)

then i tried to solve for the mass of the ice, but the Ti of water is also unknown so I could not solve it. Also, it said to ignore the heat capacity of the bucket,so i left it out of the equation completely.

Thanks for any help!

 

[Astronuc]

the Ti of water is also unknown

Think about the statement "a mixture of ice and water". Assuming the ice and water are in thermal equilibrium - what is the temperature? We have two phases solid and liquid at the same temperature, which is . . . ?