A 30.5 g sample of an alloy at 90.7�C is placed into 48.8 g water at 22.3�C in an insulated coffee cup. The heat capacity of the coffee cup (without the water) is 9.2 J/K. If the final temperature of the system is 31.1�C, what is the specific heat capacity of the alloy? (c of water is 4.184 J/g�)

Question

Jai · Answer

Recall that the heat absorbed (+) or released (-) by a substance is given by
Q = mc(T2 - T1)
where
m = mass (g)
c = specific heat capacity (J/g-K)
T2 = final temperature
T1 = initial temperature
Note that in the problem, the source of energy or heat came from the alloy. Thus we say that heat released from the alloy is absorbed by the water and cup, or:
Q,alloy + Q,water + Q,cup = 0, or
-Q,alloy = Q,water + Q,cup

Let c = specific heat capacity of alloy
Note that the final temperature is the same for all. For the cup, the given is heat capacity, not specific heat capacity (which is per mass). And the initial temperature of water & cup are the same, since they're in equilibrium before the alloy is added.
Substituting,
-(30.5)*c*(31.1 - 90.7) = (48.8)*(4.184)*(31.1-22.3) + (9.2)*(31.1-22.3)
1817.8*c = 1877.7
c = 1.03 J/g-K

Hope this helps :)