Posted by casey on .
A 10.0 grams piece of gold 90.0 Celsius was stirred into 15.0ml of water at an initial temperature of 25.0 degrees Celsius. The final temperature of the water was 26.34 degrees celsius. Calculate the specific heat of gold. The specific heat of water is 4.184 J/g degrees celsius

chemistry 
Jai,
recall that the amount of heat absorbed or released is given by:
Q = mc(T2T1)
where
m = mass (in g)
c = specific heat capacity (in J/gK)
T = temperature (in C or K)
*note: Q = (+) when heat is absorbed and () when heat is released
to get the final temp, we note that Q,released = Q,absorbed. therefore:
Q,released = Q,absorbed
m1*c1*(T2T,r1) = m2*c2*(T2T,a1)
we're solving for c1, which is the specific heat capacity of gold.
assuming that the density of water is 1g/mL, the mass of water (for a volume equal to 15 mL) is equal to 15 g. substituting,
10*c1*(26.3490) = 15*4.184*(26.3425)
636.6*c1 = 84.0984
c1 = 0.1321 J/gK
hope this helps~ :)