Answer:
1.24 kJ is required to convert 14 g of liquid from 43.5°C to 128.2°C
Explanation:
This is a typical calorimetry problem:
We have to assume, no heat is lost to sourrounding.
First of all, we need to go from 43.5°C to 97.4°C, the boiling point.
Q = Ce . m . ΔT
We replace data, 1.18° J/g . 14 g . (97.4°C - 43.5°C)
Heat for the first stage is: 890.4 Joules
Now we have to change the state, and we need the ΔH. As we do not have latent heat, we can proceed like this:
1 mol release 30.1 kJ at vaporization.
We convert the mass to moles → 14 g. 1mol/ 67.44g = 0.207 mol
0.207 mol will release (0.207 . 30.1 kJ) = 6.25 kJ
Now, we are at gaseous phase.
Q = Ce . m . ΔT → 0.792 J/g°C . 14g . (128.2°C - 97.4°C)
Q = 341.5 Joules
To determine the amount of heat, we sum all the obtained values:
890.4 Joules + 6250 Joules + 341.5 Joules = 1238.2 J
We convert to kJ → 1238.2 J . 1kJ / 1000J = 1.24 kJ
