a) 2.0 kg of molten copper at its melting point of 1083 ºC (the copper is 100% liquid) is poured into

a mold in a cooling water bath at 20 ºC. What mass of water is needed to bring the solidified
copper and water to an equilibrium temperature of 30 ºC, assuming the only heat exchange is
between the copper and water.
(Lf copper = 1.34 x 105 J/kg, ccopper = 387 J/kg ºC, cwater = 4186 J/kg ºC)

I was told to do (mass Cu*Lf Cu) + (massH2O*LfH2O*(deltaT)= 0
solve for mass H2O.

But why would they give the C of water and copper and not the Lf of water?

There are two things happening: Heat is lost as the copper solifies, and heat is lost as the copper cools. You will have to ignore any water that fizzles into vapor as that is unknown.

MassCu*LfCu + massCu*Ccu*(30-1083) + masswater*cwater*(30-20)=0
solve for mass water.

should the (mass Cu * Lf Cu) be negative because it loses heat going from a liguid to a solid?

Yes, that's correct. When the copper undergoes solidification, heat is lost from the system. Therefore, the term (mass Cu * Lf Cu) should be negative because it represents the heat lost by the copper as it changes phase from liquid to solid.