[jtknight]

I have racked my brain trying to figure out this one out. I believe that I have come up with everything except the right answer. Please *delete me*

A 5.00 g sample of aluminum pellets (specific heat capacity = 0.89 J/°Cg) and a 10.00 g sample of iron pellets (specific heat capacity = 0.45 J/°Cg) are heated to 100.0°C. The mixture of hot iron and aluminum is then dropped into 91.9 g of water at 21.1°C. Calculate the final temperature of the metal and water mixture, assuming no heat loss to the surroundings.

Appairently I am given specific heat capacity, mass, and T-final, but what do I start out solving for, and where are all the numbers get plugged into?

[Borek]

what do I start out solving for

T-final. Contrary to what you think it is not given.

and where are all the numbers get plugged into?

Heat balance. Metals lose, water gets.

[jtknight]

alright, so in order to get the heat given i add the 2 together correct?
5*.89*(100-T) + 10*.45*(100-T)
and for heat taken by water: 91.9*4.18*(T-21.1)

Do the math and come up with T?! Hopefully!

[Borek]

Looks OK. But I am known to be in error occasionally