Chapter 6, Problem 6.81E

Interpretation Introduction

Interpretation:

The total amount of heat needed to change $0.500\text{kg}$ of ice at $-10.°\text{C}$ into $0.500\text{kg}$ of steam at $120.°\text{C}$ 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 amount of energy needed to change temperature of matter is known as specific heat of matter. The amount of energy needed to change state of matter is known as heat fusion or vaporization.

Answer to Problem 6.81E

The total amount of heat needed to change $0.500\text{kg}$ of ice at $-10.°\text{C}$ into $0.500\text{kg}$ of steam at $120.°\text{C}$ is $\text{367350}\text{cal}$.

Explanation of Solution

Step 1. $\text{Ice}\left(-10.°\text{C}\right)\to \text{Ice}\left(0.°\text{C}\right)$

The formula to calculate amount of heat (in calories) needed to increase the temperature is given below as,

$\text{Heat}=\left(\text{sample}\text{mass}\right)\left(\text{specific}\text{heat}\right)\left(\text{temperature}\text{change}\right)$

Substitute the values in the above equation as follows.

$\begin{array}{rll}\text{Heat}& =& \left(\text{sample}\text{mass}\right)\left(\text{specific}\text{heat}\right)\left(\text{temperature}\text{change}\right)\\ & =& \left(0.500\times 1000\text{g}\right)\left(\frac{\text{0}\text{.51}\text{cal}}{\text{g}\text{°C}}\right)\left(0°\text{C}-\left(-10°\text{C}\right)\right)\\ & =& \left(500\text{g}\right)\left(\frac{\text{0}\text{.51}\text{cal}}{\text{g}\text{°C}}\right)\left(10°\text{C}\right)\\ & =& 2550\text{cal}\end{array}$

Step 2. $\text{Ice}\left(0.°\text{C}\right)\to \text{water}\left(0.°\text{C}\right)$

The heat of fusion is defined as the energy required for changing the state of $1\text{g}$ of 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.

$\begin{array}{rll}\text{Heat}& =& \left(\text{Heat}\text{of fusion}\right)\left(\text{sample}\text{mass}\right)\\ & =& \left(80\text{cal}/\text{g}\right)\left(0.500\times \text{1000}\text{g}\right)\\ & =& 40000\text{cal}\end{array}$

Step 3. $\text{Water}\left(0.°\text{C}\right)\to \text{water}\left(100.°\text{C}\right)$

The formula to calculate amount of heat (in calories) needed to increase the temperature is given below as,

$\text{Heat}=\left(\text{sample}\text{mass}\right)\left(\text{specific}\text{heat}\right)\left(\text{temperature}\text{change}\right)$

Substitute the values in the above equation as follows.

$\begin{array}{rll}\text{Heat}& =& \left(\text{sample}\text{mass}\right)\left(\text{specific}\text{heat}\right)\left(\text{temperature}\text{change}\right)\\ & =& \left(0.500\times 1000\text{g}\right)\left(\frac{1.0\text{cal}}{\text{g}\text{°C}}\right)\left(\text{100}°\text{C}-0°\text{C}\right)\\ & =& \left(500\text{g}\right)\left(\frac{\text{1}\text{.0}\text{cal}}{\text{g}\text{°C}}\right)\left(100°\text{C}\right)\\ & =& 50000\text{cal}\end{array}$

Step 4. $\text{Water}\left(100.°\text{C}\right)\to \text{steam}\left(100.°\text{C}\right)$

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

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

Substitute the values in the above equation as follows.

$\begin{array}{rll}\text{Heat}& =& \left(\text{Heat}\text{of vaporization}\right)\left(\text{sample}\text{mass}\right)\\ & =& \left(540\text{cal}/\text{g}\right)\left(0.500\times \text{1000}\text{g}\right)\\ & =& 270000\text{cal}\end{array}$

Step 5. $\text{Steam}\left(100.°\text{C}\right)\to \text{steam}\left(120.°\text{C}\right)$

The formula to calculate amount of heat (in calories) needed to increase the temperature is given below as,

$\text{Heat}=\left(\text{sample}\text{mass}\right)\left(\text{specific}\text{heat}\right)\left(\text{temperature}\text{change}\right)$

Substitute the values in the above equation as follows.

$\begin{array}{rll}\text{Heat}& =& \left(\text{sample}\text{mass}\right)\left(\text{specific}\text{heat}\right)\left(\text{temperature}\text{change}\right)\\ & =& \left(0.500\times 1000\text{g}\right)\left(\frac{\text{0}\text{.48}\text{cal}}{\text{g}\text{°C}}\right)\left(120°\text{C}-100°\text{C}\right)\\ & =& \left(500\text{g}\right)\left(\frac{\text{0}\text{.48}\text{cal}}{\text{g}\text{°C}}\right)\left(20°\text{C}\right)\\ & =& 4800\text{cal}\end{array}$

The total amount of heat needed to change $0.500\text{kg}$ of ice at $-10.°\text{C}$ into $0.500\text{kg}$ of steam at $120.°\text{C}$ is calculated as,

$\begin{array}{rll}\text{Heat}& =& 2550\text{cal}+40000\text{cal}+50000\text{cal}+270000\text{cal}+4800\text{cal}\\ & =& \text{367350}\text{cal}\end{array}$

Conclusion

The total amount of heat needed to change $0.500\text{kg}$ of ice at $-10.°\text{C}$ into $0.500\text{kg}$ of steam at $120.°\text{C}$ is $\text{367350}\text{cal}$.