20.0 g of ammonium nitrate (NH4NO3) is dissolved in 125 g of water in a coffee-cup calorimeter, the temperature falls from 296.5 K to 286.4 K. Find the value of q for the calorimeter.

Answer 1

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/g-K)
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/g-K.
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