Answer
During this freezing, 2800 cal of heat were absorbed.
Work Step by Step
1. Identify the objective.
- Find the number of calories that are necessary to freeze 35 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 $\frac{1g}{80.cal}$
3. Using the conversion factor, calculate the necessary heat:
$35g \times \frac{80.cal}{1g} = 2800 cal$
4. Adjust the number to the correct number of significant figures.
- The used number that has the fewest number of significant figures is "35", with 2. Therefore, the result of the multiplication must have 2 SFs.
2800 cal = 2800 cal
5. Indicate whether heat was absorbed or released.
- During the freezing process, heat is absorbed.
You can check this information on page 76.
