Q. 3.46 - An Introduction to General, Organic, and Biological Chemistry

Question

Calculate the heat change at $0°$. for each of the following. and indicate whether heat was absorbed/released:

a. calories to freeze$35 \mathrm{g}$ of water

h. joules to freeze$275 \mathrm{g}$ of water

c. kilocalories to melt$140 \mathrm{g}$ of ice

d. kilojoules to melt$5.00 \mathrm{g}$ of ice

Step-by-Step Solution

Answer

As a result, the heat change is $2.8\times {10}^3\mathrm{cal}$As a result, the heat was released during the freezing process.

As a result, the heat change is$9.19\times {10}^4\mathrm{J}$. As a result, the heat was released during the freezing process.

As a result, the heat change is $11.2\mathrm{kca}$As a result, the heat was released during the freezing process.

As a result, the heat change  is $1.67\mathrm{kJ}$As a result, the heat was absorbed during the melting process.

Step 1: Given data

Calculate the heat change at $0°C$. for each of the following. and indicate whether heat was absorbed/released

Step 2: The heat was released during the process of freezing (part a)

(a)

The mass of water  is $35g$and the heat of fusion of water in calorie per gram is$80cal/g$ .Now, use the following formula to calculate the change in temperature:

Heat change $=$ mass $\times$ heat of fusion

$=35\mathrm{g}\times \frac{80\mathrm{cal}}{\mathrm{g}}$

$=2.8\times {10}^3\mathrm{cal}$

Therefore, the heat change is.$2.8\times {10}^3\mathrm{cal}$As a result, the heat was released during the freezing process.

Step 3:The mass of water (part b)

(b)

The mass of water is$275g$ and the heat of fusion of water in joule per gram is$334J/g$.

Now, use the following formula to calculate the change in temperature:

Heat change$=$ mass$\times$ heat of fusion

$=275\mathrm{g}\times \frac{334\mathrm{J}}{\mathrm{g}}$

$=9.19\times {10}^4\mathrm{J}$

Therefore, the heat change$=9.19\times {10}^4\mathrm{J}$. As a result, the heat was released during the freezing process.

Step 4:The mass of ice(part c)

(c)

The mass of ice is $140g$ and the heat of fusion of ice in calorie per gram is$80\mathrm{cal}/\mathrm{g}$ .Now, use the following formula to calculate the change in temperature:

Heat change$=$ mass $\times$ heat of fusion

$=140\mathrm{g}\times \frac{80\mathrm{cal}}{\mathrm{g}}\times \frac{0.001\mathrm{kcal}}{1\mathrm{cal}}$

$=11.2\mathrm{kcal}$

Therefore, the heat change is$=11.2\mathrm{kcal}$ As a result, the heat was absorbed during the melting process.

Step 5: The heat change (part d)

(d)

The mass of ice is $5.00 \mathrm{g}$and the heat of fusion of ice in joule per gram is $334 \mathrm{J}/\mathrm{g}.$Now, use the following formula to calculate the change in temperature:

Heat change $=$mass $\times$ heat of fusion

$=5.00\mathrm{g}\times \frac{334\mathrm{J}}{\mathrm{g}}\times \frac{1\mathrm{kJ}}{1000\mathrm{J}}$

$=1.67\mathrm{kJ}$

Therefore, the heat change is $=1.67\mathrm{kJ}$. As a result, the heat was absorbed during the melting process.