Chapter 6, Problem 6.78E

Interpretation Introduction

(a)

Interpretation:

The heat that could be stored by melting $1000\text{kg}$ of calcium chloride $\left({\text{CaCl}}_{\text{2}}\cdot {\text{6H}}_{\text{2}}\text{O}\right)$ having $40.7\text{cal}/\text{g}$ heat of fusion is to be calculated.

Concept introduction:

When temperature is changed or the state of matter is changed the energy is either absorbed or released. The energy required to change temperature of matter is known as specific heat of matter. The energy required to change a state of matter is known as heat of fusion or vaporization.

Answer to Problem 6.78E

The heat that could be stored by melting $1000\text{kg}$ of calcium chloride $\left({\text{CaCl}}_{\text{2}}\cdot {\text{6H}}_{\text{2}}\text{O}\right)$ having $40.7\text{cal}/\text{g}$ heat of fusion is $4.07\times {\text{10}}^7\text{cal}$.

Explanation of Solution

The heat of fusion is defined as the energy required for changing the state of $1\text{g}$ substance. Thus, the energy that could be stored by melting a solid is given by the formula written below.

$\text{Heat}=\left(\text{Heat}\text{of fusion}\right)\left(\text{sample}\text{mass}\right)$

Substitute the values in the above equation as follows.

$\begin{array}{rll}\text{Heat}& =& \left(\text{Heat}\text{of fusion}\right)\left(\text{sample}\text{mass}\right)\\ & =& \left(40.7\text{cal}/\text{g}\right)\left(1000\times \text{1000}\text{g}\right)\\ & =& 4.07\times {\text{10}}^7\text{cal}\end{array}$

Conclusion

The heat that could be stored by melting $1000\text{kg}$ of calcium chloride $\left({\text{CaCl}}_{\text{2}}\cdot {\text{6H}}_{\text{2}}\text{O}\right)$ having $40.7\text{cal}/\text{g}$ heat of fusion is $4.07\times {\text{10}}^7\text{cal}$.

Interpretation Introduction

(b)

Interpretation:

The heat that could be stored by melting $1000\text{kg}$ of Lithium nitrate $\left({\text{LiNO}}_3\cdot 3{\text{H}}_{\text{2}}\text{O}\right)$ having $70.7\text{cal}/\text{g}$ heat of fusion is to be calculated.

Concept introduction:

When temperature is changed or the state of matter is changed the energy is either absorbed or released. The energy required to change temperature of matter is known as specific heat of matter. The energy required to change a state of matter is known as heat of fusion or vaporization.

Answer to Problem 6.78E

The heat that could be stored by melting $1000\text{kg}$ of Lithium nitrate $\left({\text{LiNO}}_3\cdot 3{\text{H}}_{\text{2}}\text{O}\right)$ having $70.7\text{cal}/\text{g}$ heat of fusion is $7.07\times {\text{10}}^7\text{cal}$.

Explanation of Solution

The heat of fusion is defined as the energy required for changing the state of $1\text{g}$ substance. Thus, the energy that could be stored by melting a solid is given by the formula written below.

$\text{Heat}=\left(\text{Heat}\text{of fusion}\right)\left(\text{sample}\text{mass}\right)$

Substitute the values in the above equation as follows.

$\begin{array}{rll}\text{Heat}& =& \left(\text{Heat}\text{of fusion}\right)\left(\text{sample}\text{mass}\right)\\ & =& \left(70.7\text{cal}/\text{g}\right)\left(1000\times \text{1000}\text{g}\right)\\ & =& 7.07\times {\text{10}}^7\text{cal}\end{array}$

Conclusion

The heat that could be stored by melting $1000\text{kg}$ of Lithium nitrate $\left({\text{LiNO}}_3\cdot 3{\text{H}}_{\text{2}}\text{O}\right)$ having $70.7\text{cal}/\text{g}$ heat of fusion is $7.07\times {\text{10}}^7\text{cal}$.

Interpretation Introduction

(c)

Interpretation:

The heat that could be stored by melting $1000\text{kg}$ of Sodium sulfate $\left({\text{Na}}_2{\text{SO}}_4\cdot 10{\text{H}}_{\text{2}}\text{O}\right)$ having $57.1\text{cal}/\text{g}$ heat of fusion is to be calculated.

Concept introduction:

When temperature is changed or the state of matter is changed the energy is either absorbed or released. The energy required to change temperature of matter is known as specific heat of matter. The energy required to change a state of matter is known as heat of fusion or vaporization.

Answer to Problem 6.78E

The heat that could be stored by melting $1000\text{kg}$ of Sodium sulfate $\left({\text{Na}}_2{\text{SO}}_4\cdot 10{\text{H}}_{\text{2}}\text{O}\right)$ having $57.1\text{cal}/\text{g}$ heat of fusion is $5.71\times {\text{10}}^7\text{cal}$.

Explanation of Solution

The heat of fusion is defined as the energy required for changing the state of $1\text{g}$ substance. Thus, the energy that could be stored by melting a solid is given by the formula written below.

$\text{Heat}=\left(\text{Heat}\text{of fusion}\right)\left(\text{sample}\text{mass}\right)$

Substitute the values in the above equation as follows.

$\begin{array}{rll}\text{Heat}& =& \left(\text{Heat}\text{of fusion}\right)\left(\text{sample}\text{mass}\right)\\ & =& \left(57.1\text{cal}/\text{g}\right)\left(1000\times \text{1000}\text{g}\right)\\ & =& 5.71\times {\text{10}}^7\text{cal}\end{array}$

Conclusion

The heat that could be stored by melting $1000\text{kg}$ of Sodium sulfate $\left({\text{Na}}_2{\text{SO}}_4\cdot 10{\text{H}}_{\text{2}}\text{O}\right)$ having $57.1\text{cal}/\text{g}$ heat of fusion is $5.71\times {\text{10}}^7\text{cal}$.