http://www.ck12.org Chapter 17. Thermochemistry
- Calculation of the energy needed to raise the temperature of the ice from -12°C to 0°C
- Calculation of the energy needed for the transition from solid ice to liquid water (calculated in the previous
example problem) - Calculation of the energy needed to raise the temperature of the liquid water from 0°C to 25°C
The portions that require calculating the energy associated with temperature changes will make use of the following
equation:
∆H=m×c×∆T
First, calculate the energy associated with the temperature change for ice:
∆H= (180 g)(^2 g.^09 ·◦CJ)( 12 ◦C) = 4514 .4 J
Second, recall from the previous problem that melting 180 grams of ice requires 60 kJ (60,000 J) of energy.
Third, calculate the energy needed to change the temperature of the liquid water. Again, note that ice and liquid
water have different specific heat values.
∆H= (180 g)(^4 g.^18 ·◦CJ)( 25 ◦C) =18810 J
Finally, add all of these energetic requirements together to find the energy needed for the entire change. Make sure
that all values have the same units (J or kJ, but not both).
∆H= 4 , 514 .4 J+ 60 ,000 J+ 18 ,810 J= 83 , 324 .4 J
Rounded to the correct number of significant figures, this process would require the input of 83,000 J (83 kJ) of
energy.
Note that the enthalpy change is positive, indicating that we need to put energy into the system in order to achieve the
desired transition. When going “downhill” from vapor to liquid or liquid to solid, we need to remember to change
the sign for∆H. If we were to do the reverse process, cooling liquid water at 25°C to ice at -12°C,∆H would be -83
kJ, indicating that 83 kJ of energy would be removed from the system during this transition.
Heat of Solution
If you were to prepare a large solution of sulfuric acid in water, it would be wise to cool the container in which the
mixing is taking place. The addition of a strong acid to water releases a great deal of heat (which is why it is better
to add the acid to the water and not the water to the acid). Theheat of solutionfor a substance tells us how much
energy is absorbed or released when one mole of the substance is fully dissolved in a specific solvent. We will only
consider heat of solution values for which water is the solvent, but other solvents would have different values for a
given solute. The heats of solution for several common solutes are listed in theTable17.3. In each case, assume
that the solute is added to a large excess of water, so the solute dissolves completely.
TABLE17.3:
Material Heat of Solution (kJ/mol)
H 2 SO 4 (l) -96.2
MgSO 4 (s) -91.2
CaCl 2 (s) -82.9
KOH(s) -56
NaOH(s) -44.3
NaCl(s) 3.9
NaHCO 3 (s) 16.7
KNO 3 (s) 35