help! Determine the specific heat of the unknown sample?

Question

An aluminum calorimeter with a mass of 105 g contains 240 g of water. The calorimeter and water are in thermal equilibrium at 14.5 degrees C. Two metallic blocks are placed into the water. One is a 56.5-g piece of copper at 67 degrees Celsius. The other block has a mass of 333 g and is originally at a temperature of 124 degrees Celsius. The entire system stabilizes at a final temperature of 25 degrees
Celsius.
Determine the specific heat of the unknown sample. Take the specific heat of water to be 1 cal/kg*degree C.

specific heat of aluminum = .216 cal/g*degrees C
not sure if you need it though...

Answer 1

dQ(water) + dQ(cupper) + dQ(aluminum) + dQ(unknown or x)=0

dQ=Cp m (T2-T1)
dQ -change in heat
T2 - final temperature
T1 - initial temperature
Cp - specific heat of the substance
m - mass of the substance

Note there is a problem. The unknown is at 124 C. This is 24 degrees above boiling point of water. This can be a small contributor to an error since mass of water is reduced and we have to incorporate the change in phase going from liquid to gas (latent heat of evaporation).

Lets pretend the temperature of the unknown will not cause any of the water to boil. Then we can continue.

T2 = 25C
Then
dQ(water) + dQ(copper) + dQ(aluminum) + dQ(unknown or x)=0

becomes

1.0 x .240 x(25 -14.5) + 0.0924 x 0.0565 x (25 - 67) + 0.215 x 0.105 (25 -14.5) + Cp(of x) 0.333(25-124)=0

Now just solve for Cp

Cp(of x)=[ 1.0 x .240 x(25 -14.5) + 0.0924 x 0.0565 x (25 - 67) + 0.215 x 0.105 (25 -14.5)]/ [0.333(124 - 25)]