Question 11.3: Calculating the Amount of heat of a Temperature Change How m...
Calculating the Amount of heat of a Temperature Change
How much heat is required to convert 50.0 g of water at 25 °C to steam at 150 °C? The boiling point of water is 100 °C and C_{m} [H_{2}O(l) ] = 75.4 J/(mol . °C),∆H_{vap} = 40.67 kJ/mol, Cm [H_{2}(g)] = 33.6 J/(mol . °C).
IDENTIFY
| Known | Unknown |
| Temperature change (25 °C to 150 °C) | Heat (q in kJ) |
| Amount of water (50.0 g) | |
| Boiling point 100 °C | |
| ∆H_{vap} = 40.67 kJ/mo | |
| C_{m} [H_{2}O(l)] = 75.4 J/(mol . °C) | |
| C_{m}[H_{2}O(g)] = 33.6 J/(mol .°C) | |
STRATEGY
There are three separate steps involved in heating the sample. Step 1 is heating the liquid water to its boiling point, step 2 is vaporizing the liquid water, and step 3 is heating the water vapor. First convert from grams to moles since the molar heat capacities are given and then calculate q for each step.
First find the number of moles of water:
50.0 g H_{2}O\times \frac{1 mol H_{2}O}{18.0 g H_{2}O}=2.78 mol H_{2}O
Step 1. Heating liquid H_{2}O from 25 °C to 100 °C:
q = C_{m} \times Moles of a substance \times ∆T =(75.4\frac{J}{mol .°C} )(2.78 mol)(75 °C)=1.57\times 10^{4} J=15.7 kJStep 2. Vaporizing liquid H_{2}O:
q = ∆H_{vap} \times Moles of a substance = (40.67 kJ/mol) (2.78 mol)= 113.0 kJStep 3. Heating gaseous H_{2}O from 100 °C to 150 °C:
q = C_{m} \times Moles of a substance \times ∆T =(33.6\frac{J}{mol.°C} )(2.78 mol)(50 °C)=4.67\times 10^{3} J=4.67 kJThe total heat required is a sum of all three steps:
15.7 kJ + 113.0 kJ + 4.67 kJ = 133 kJ
CHECK
The amount of heat is large and positive, which is reasonable for a phase change. Also the magnitude of the heat for each step is reasonable. The phase change from liquid to vapor requires the most heat because strong hydrogen bonds between water molecules must be broken. Raising the temperature of liquid water requires more heat than raising the temperature of steam because the molar heat capacity and the temperature change are larger for the liquid.