Answer
5200 calories were absorbed during the process.
Work Step by Step
1. Identify the objective.
- Find the number of calories that are necessary to freeze 225 g of water at 0$^{\circ}$.
2. Find the conversion factors.
- To convert the mass in g to calories in a freezing process, we can use the Heat of Fusion for water.
Page 76: $\frac{80. cal}{1g}$ and $\frac{1g}{80.cal}$
And: $1 kcal = 1000 cal$
3. Using the conversion factor, calculate the necessary heat:
$225g \times \frac{80. cal}{1g} \times \frac{1kcal}{1000cal} = 18 kcal$
4. Adjust the number to the correct number of significant figures.
- The used number that has the fewest number of significant figures is "80.", with 2. Therefore, the result of the multiplication must have 2 SFs.
18 kcal = 18 kcal.
5. Indicate whether heat was absorbed or released.
- During the freezing process, heat is released.
You can check this information on page 76.
