Answer
During this vaporization process, 24 kcal of heat were absorbed.
Work Step by Step
1. Identify the objective.
- Find the number of kilocalories that are necessary to vaporize 44 g of water at 100$^{\circ}$.
2. Find the conversion factors.
- To convert the mass in g to calories in a vaporization process, we can use the Heat of Vaporization for water.
Page 79: $\frac{540cal}{1g}$ and $\frac{1g}{540cal}$
And: $1kcal = 1000cal$
3. Using the conversion factor, calculate the necessary heat:
$44g \times \frac{540cal}{1g} \times \frac{1kcal}{1000cal}= 23.76 kcal$
4. Adjust the number to the correct number of significant figures.
- The used number that has the fewest number of significant figures is "44", with 2. Therefore, the result of the multiplication must have 2 SFs.
23.76 kcal = 24 kcal
5. Indicate whether heat was absorbed or released.
- During the vaporization process, heat is absorbed.