When 0.100 g of calcium oxide is added to 125 g of water at 23.6°C in a coffee cup calorimeter, this reaction occurs:
CaO (s) + H2O(l) -> Ca(OH)2 (aq)
∆H = -81.9kJ/mol
If the specific heat capacity of water is 4.184 J g-1 °C-1, what is the final temperature of the solution?
To find the final temperature of the solution, we can make use of the principle of conservation of energy. The energy released by the exothermic reaction (the heat released) will be absorbed by the surrounding water. We can set up an equation using the heat transfer equation:
q_water + q_calorimeter = 0
The heat absorbed by the water, q_water, can be calculated using the formula:
q_water = m_water * C_water * ∆T_water
where:
- m_water is the mass of water (125 g),
- C_water is the specific heat capacity of water (4.184 J g^(-1) °C^(-1)),
- ∆T_water is the change in temperature of water (final temperature - initial temperature).
Since the calcium oxide is in excess and completely reacts with water, we can assume that the heat absorbed by the calorimeter, q_calorimeter, is negligible.
Now, let's rearrange the equation and solve for ∆T_water:
q_water = -q_calorimeter
m_water * C_water * ∆T_water = -81.9 kJ/mol
To cancel out the units, we can convert the heat released (∆H) from kJ/mol to J/g:
∆H = -81.9 kJ/mol = -81900 J/0.100 g = -819000 J/g
Substituting the values into the equation, we get:
125 g * 4.184 J g^(-1) °C^(-1) * ∆T_water = -819000 J/g
Now, we can calculate ∆T_water by rearranging the formula:
∆T_water = (-819000 J/g) / (125 g * 4.184 J g^(-1) °C^(-1))
Evaluating this calculation will give us the change in temperature of the water (∆T_water).
Finally, to find the final temperature of the solution, we need to add ∆T_water to the initial temperature.
figure the moles of CaO in 100 g. Multiply that by -81.8 kJ/mol. Notice the negative sign, that means heat is not absorbed.
that is the heat given off.
notice the amount of CaO is trivial compared to the water, so assume all of the Water does not react, and there is 100g of water being heated.
finally, the sum of heats ABSORBED is zero...
100*cwater*(Tf-23.6)+heatabove=0
solve for Tf