Calorimetry Help: Calculating Final Temp of Iron & Aluminum in Water

[viper2308]

Calorimetry HELP!

Worked on this problem for an hour and cannot figure it out. Please help.

A 5.00 g sample of aluminum pellets (specific heat capacity = 0.89 J/°C·g) and a 10.00 g sample of iron pellets (specific heat capacity = 0.45 J/°C·g) are heated to 100.0°C. The mixture of hot iron and aluminum is then dropped into 90.1 g of water at 21.0°C. Calculate the final temperature of the metal and water mixture, assuming no heat loss to the surroundings.

 

[bomba923]

Hint: q = mC_p \Delta T

 

[AbedeuS]

Yeah, its a tough one, because you must work with adding specific heat capacities (not heat capacities based off mass, but the specific ones) to find the final temperature, this would be like adding Cv(aluminium) . m(aluminium) + Cv(Water) . m(water) together to find a final pseudo specific heat capacity.

P.S. I really need to learn how to use those math symbols >,<

 

[eli64]

remember too that q from both metals will be absorbed by the water. you can equate the heat liberated from the metals to the heat absorbed by the water

it is just a long algebra problem,

having the equation like this might help

q = Cp* m *(Tf-Ti)

you know everything except Tf ; equate heat liberated to heat absorbed