Posted by **casey** on Thursday, October 20, 2011 at 10:53pm.

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**, Thursday, October 20, 2011 at 11:04pm
recall that the amount of heat absorbed or released is given by:

Q = mc(T2-T1)

where

m = mass (in g)

c = specific heat capacity (in J/g-K)

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*(T2-T,r1) = m2*c2*(T2-T,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*(26.34-90) = 15*4.184*(26.34-25)

636.6*c1 = 84.0984

c1 = 0.1321 J/g-K

hope this helps~ :)

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