4. A calorimeter contains 500 g of water at 25°C. You place a cold pack containing 100 g of crystalline ammonium nitrate inside the calorimeter. When the ammonium nitrate finishes dissolving, the temperature of the water inside the calorimeter is 9.3°C. The specific heat of water is 4.18 J/g–°C. What is the enthalpy of fusion (ΔHf) of the ammonium nitrate? (Show your work.) Where necessary, use q = mHf.
this one I definitely am confused on Dr B, i tried remembering but nothing
I'm curious. Do you mean delta H formation or delta H fusion or could you mean delta H solution
it would probably be fusion, right.... based on the rest of the question? since it's asking for specific heat
OK, so I'm guessing that you're guessing as to the meaning. If you want the delta H solution here is how to go about it.
q = mass H2O x specific heat H2O x (Tfinal-Tinitial)
q = 500 x 4.18 x (9.3-25) = -32,813 joules and that is the result 2ith 100 g NH4NO3. mols NH4NO3 = 100 g x (1 mol/80g) = 1.25
Then 32,813/1.25 = 26,250 or 26.2 kJ/mol.
To find the enthalpy of fusion (ΔHf) of ammonium nitrate, we need to use the equation:
q = m × ΔHf
Where:
q = heat absorbed or released by the substance (in this case, water)
m = mass of the substance (in this case, water)
ΔHf = enthalpy of fusion
Now, let's calculate the heat absorbed by the water (q) using the formula:
q = m × c × ΔT
Where:
m = mass of the water = 500 g
c = specific heat of water = 4.18 J/g-°C
ΔT = change in temperature of the water = (9.3°C - 25°C)
ΔT = -15.7°C (negative sign indicates a decrease in temperature)
Plugging the values into the equation, we get:
q = 500 g × 4.18 J/g-°C × (-15.7°C)
q = - 39,215 J
Since the water absorbs the heat released by the dissolving of ammonium nitrate, the heat released by the ammonium nitrate can be considered equal in magnitude but opposite in sign (positive) to the heat absorbed by the water. Therefore, q for the ammonium nitrate is + 39,215 J.
We are given the mass of the ammonium nitrate as 100 g. Now, we can calculate the enthalpy of fusion (ΔHf) of the ammonium nitrate using the equation:
q = m × ΔHf
Rearranging the equation, we get:
ΔHf = q / m
Plugging in the values, we get:
ΔHf = 39,215 J / 100 g
ΔHf = 392.15 J/g
Therefore, the enthalpy of fusion (ΔHf) of ammonium nitrate is 392.15 J/g.