Answer
During this melting process, 11 kcal of heat were absorbed.
Work Step by Step
1. Identify the objective.
- Find the number of kilocalories that are necessary to melt 140 g of ice at 0$^{\circ}$.
2. Find the conversion factors.
- To convert the mass in g to calories in a melting 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:
$140g \times \frac{80.cal}{1g} \times \frac{1kcal}{1000cal} = 11.2 kcal$
4. Adjust the number to the correct number of significant figures.
- The used number that has the fewest number of significant figures is "140", with 2. Therefore, the result of the multiplication must have 2 SFs.
11.2 kJ = 11 kJ
5. Indicate whether heat was absorbed or released.
- During the melting process, heat is absorbed.
You can check this information on page 76.
