Q. 3.45 - An Introduction to General, Organic, and Biological Chemistry

Question

Calculate the heat change at $0\times \mathrm{C}$ for each of the following. and indicate whether heat was hashed/released:

a. calories to melt $65 \mathrm{g}$of ice

b. joules to melt $17.0 \mathrm{g}$ of ice

c. kilocalories to freeze$225 \mathrm{g}$ of water

d. kilojoules to freeze$50.0 \mathrm{g}$ of water

Step-by-Step Solution

Answer

As a result, the calories absorbed to melt$65g$  of ice are $5200cal$.

As a result, the calories absorbed to melt $17.0g$ office at $5680J$

As a result, the calories absorbed to melt $255g$ of water are$18.0kcal$ .

As a result, the calories absorbed to melt $50.0g$ of water are $16.7 J$

Step 1:Given data(part a)

(a) Energy is defined as the ability to perform work. Thermal energy, also known as heat, is linked to particle motion. When a substance is heated, heat is absorbed and the temperature rises due to particle movement..

Calculate the calories needed to melt $65 \mathrm{g}$ of ice at $0{}^{\circ }\mathrm{C}$. The heat equation is as follows:

$\text{ Heat }=\text{ mass }\times H_{\mathrm{f}}$

Step 2:Determine the energy in joules associated in melting (part b)

a)

The heat of the fusion for water is.$80.0\mathrm{cal}/\mathrm{g}$

The conversion factor needed to convert grams to calories is:

$1 \mathrm{g}\text{ of }{\mathrm{H}}_2\mathrm{O}(s\to l)=80.0\mathrm{cal}/\mathrm{g}$

Hence,

$\text{ Conversion factor }=\frac{80.0\mathrm{cal}}{1 {\mathrm{gH}}_2\mathrm{O}}$

Determine the energy required to melt$65g$ of ice at$0°C$ by using the following equation:

$=65{\mathrm{gH}}_2\mathrm{O}\times \frac{80.0\mathrm{cal}}{1{\mathrm{gH}}_2\mathrm{O}}$

$=5200cal$

When ice absorbs heat, the particle motion increases, and the ice melts into liquid water. When ice melts, its state changes from solid to liquid, and heat is absorbed as a result.

As a result, the calories absorbed to melt$65g$ of ice are $5200cal$.

Step 3:Given data (part b)

(b) Determine the energy in joules associated in melting$17.0g$ of ice at $0°C$. The heat equation is as follows:

Heat $=mass\times H$

The heat of the fusion for water is $334J/g$.

To convert grams to joules, use the following conversion factor::

$1 {\mathrm{gofH}}_2\mathrm{O}(s\to l)=334 \mathrm{J}$

As a result, Conversion factor$=\frac{334 \mathrm{J}}{1 {\mathrm{gH}}_2\mathrm{O}}$

Step 4:Determine the kcal to freeze (part b)

b)Determine the joules required to melt $17.0g$of ice at$0°C$ by using the following equation:

Heat $=$mass$\times H_{\mathrm{f}}$

$=17.0{\mathrm{gH}}_2\mathrm{O}\times \frac{334\mathrm{J}}{1{\mathrm{gH}}_2\mathrm{O}}$

$=5678J$

$\approx 5680J$

Because of the increase in particle motion, when heat is absorbed by ice, it is converted into liquid water.. Hence,$5680J$ are absorbed.

As a result, Conversion factor$17.0g$ office at $0°C$ are $5680J$

Step 5:Given data (part c)

(c) Determine the kilocalories to freeze$225g$ of water at $0°C$. The heat equation is as follows:

Heat $=$mass$\times H_{\mathrm{f}}$

The heat of the fusion for water is $80cal/g$. The conversion factor to convert grams to Cal is as follows:

${\mathrm{H}}_2\mathrm{O}(l\to s)=80.0\mathrm{cal}$

Hence, we have

Conversion factor $=\frac{80.0\mathrm{cal}}{1 {\mathrm{gH}}_2\mathrm{O}}$

Step 6:Determine kcal to freeze(part c)

c)Determine the kcal to freeze$225g$ of water at $0°C$ by using the following equation:

Heat$=$ mass $\times H_{\mathrm{f}}$

We have obtained by substituting the values in the equation

$\text{ Heat }=225{\mathrm{gH}}_2\mathrm{O}\times \frac{80.0\mathrm{eat}}{1{\mathrm{gH}}_2\mathrm{O}}\times \frac{1\mathrm{kcal}}{{10}^3\mathrm{eat}}$

$=18.0kcal$

When liquid water turns into ice, heat is released. The procedure is known as freezing. As a result, heat is removed in order to freeze.$225g$ water. Therefore, the kilocalories released to freeze $225g$ of water are $18.0kcal$.

Step 7:Given data(part d)

(d) Determine the energy in kilojoules associated in freezing$50.0g$ of ice at $0°C$. The heat equation is as follows:

Heat $=$mass$\times H_{\mathrm{f}}$

The heat of the fusion for water is$334J/g$.

The conversion factor to convert grams to joules is as follows:

$1 \mathrm{g}\text{ of }{\mathrm{H}}_2\mathrm{O}(s\to l)=334 \mathrm{J}$

Step 8:Determine the joules required to freeze (part d)

d)Hence we have

Conversion factor $=\frac{334 \mathrm{J}}{1 {\mathrm{gH}}_2\mathrm{O}}$

The conversion factor to convert joules to kilojoules is as follows:

 joules$1 \mathrm{kJ}={10}^3$

Hence, Conversion factor

$=\frac{1 \mathrm{kJ}}{{10}^3\text{ joules }}$

Determine the joules required to freeze $50.0g$ of ice at $0°C$ by using the following equation:

$\text{ Heat }=\text{ mass }\times H_f$

$=50.0{\mathrm{gH}}_2\mathrm{O}\times \frac{334+}{1{\mathrm{gH}}_2\mathrm{O}}\times \frac{1\mathrm{kJ}}{{10}^3\mathrm{f}}$

$=16.7J$

When liquid water turns into ice, heat is released.. The procedure is known as freezing. As a result, heat is removed in order to freeze. $50.0g$ of water.

 Therefore, the kilocalories released to freeze$50.0g$ of water are $16.7J$