Answer
During this vaporization process, 11300 kJ of heat were absorbed.
Work Step by Step
1. Identify the objective.
- Find the number of kilojoules that are necessary to vaporize 5.00 kg of water at 100$^{\circ}$.
2. Find the conversion factors.
- To convert the mass in g to joules in a vaporization process, we can use the Heat of Vaporization for water.
Page 79: $\frac{2260J}{1g}$ and $\frac{1g}{2260J}$
$1 kJ = 1000 J$
$1 kg = 1000 g$
3. Using the conversion factors, calculate the necessary heat:
$5.00 kg \times \frac{1000g}{1kg} \times \frac{2260 J}{1g} \times \frac{1kJ}{1000J} = 11300 kJ$
4. Adjust the number to the correct number of significant figures.
- The used number that has the fewest number of significant figures is "5.00", with 3. Therefore, the result of the multiplication must have 3 SFs.
11300 kJ = 11300 kJ
5. Indicate whether heat was absorbed or released.
- During the vaporization process, heat is absorbed.