91Ó°ÊÓ

(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 $0{}^{∘}\mathrm{C}$. The heat equation is as follows:

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

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:

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

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:

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

$=5200cal$

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 $5200cal$.

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

Heat $=mass×H$

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

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

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

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

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

Heat$=$mass$×H_{\mathrm{f}}$

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

$=5678J$

$≈5680J$

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 $5680J$

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

Heat$=$mass$×H_{\mathrm{f}}$

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

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

Hence, we have

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

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

Heat$=$mass $×H_{\mathrm{f}}$

We have obtained by substituting the values in the equation

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

$=18.0kcal$

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 $18.0kcal$.

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

Heat $=$mass$×H_{\mathrm{f}}$

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

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

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

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:

$\text{Heat}=\text{mass}×H_f$

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

$=16.7J$

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 $16.7J$