Final temperature when mixing water, aluminum and copper

[Bri]

A combination of 0.250 kg of water at 20*C, 0.400 kg of aluminum at 26.0*C, and 0.100 kg of copper at 100*C is mixed in an insulated container and allowed to come to thermal equilibrium. Ignore any energy transfer to or from the container and determine the final temperature of the mixture.

I know to use Q=mc(delta)T and that the energy lost should equal the energy gained, but how do I set up the equation for problems like this?
One of the things that I don't know how to figure out is whether the aluminum will be losing energy to the water or gaining it from the copper. So I don't know how to set this up.

Thanks.

 

[dextercioby]

HINT:The total heat transferred must be zero (the absorbed one is positive and the given one is negative,when u add them,they will yield zero).

Daniel.

 

[Doc Al]

I know to use Q=mc(delta)T and that the energy lost should equal the energy gained, but how do I set up the equation for problems like this?
One of the things that I don't know how to figure out is whether the aluminum will be losing energy to the water or gaining it from the copper. So I don't know how to set this up.

Let the equations do the work for you. Each substance (water, copper, aluminum) will have a heat transfer: $\Delta Q_1 = m_1c_1 (T_f - T_i)$. The total $\Delta Q = \Delta Q_{water} + \Delta Q_{aluminum} + \Delta Q_{copper}$ equals zero and the final temp is the same for all.