chemistry
posted by Sona .
20.0 g of ammonium nitrate (NH4NO3) is dissolved in 125 g of water in a coffeecup calorimeter, the temperature falls from 296.5 K to 286.4 K. Find the value of q for the calorimeter.

When NH4NO3 is dissolved in water, it dissociates into its ions, followed by a temperature drop:
NH4NO3 > NH4+ + NO3
(though it's not really that relevant though ^^;)
Anyway, the heat gained or lost can be calculated using the formula
Q = mc(T2  T1)
where
m = mass (grams)
c = specific heat capacity (J/gK)
T = temperature (K)
Here, we'll use an assumption that the specific heat of the solution is equal to that of water, which is c = 4.184 J/gK.
The total mass of solution is 20 + 125 = 145 grams.
Substituting,
Q = (145)*(4.184)*(286.4  296.5)
Q = 6127.47 J
The Q negative because heat is released in the dissociation.
Check the significant figures.
Hope this helps :3
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