A 24.95 g aluminum block is warmed to 65.15 °C and plugged into an insulated beaker containing 99.41 g water i?
A 24.95 g aluminum block is warmed to 65.15 °C and plugged into an insulated beaker containing 99.41 g water initially at 18.29 °C. The aluminum and the water are allowed to come to thermal equilibrium. Assuming that no heat is lost, what is the final temperature of the water and aluminum?
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Some discussion and several solved sample problems:
http://chemteam.info/Thermochem/MixingMetal&Water....
For your problem:
(24.95 g) (65.15 - x) (____) = (99.41 g) (x - 18.29) (4.184 J/g C)
Look up the specific heat of aluminum and put it where the (____) is. IIRC, the value is 0.901 J/g C.
Then, solve for x, the final temperature of the system.