Chapter 12, Problem 60A

To determine

The thermal energy added to $10.0\text{g}$ of ice at $-20.0°\text{C}$ to convert it to water vapor at $120.0°\text{C}$ .

Answer to Problem 60A

  $3.094\times {10}^4\text{J}$

Explanation of Solution

Given:

The mass of the ice, $m=10.0\text{g = 0}\text{.01}\text{kg}$

Initial temperature of ice, $T_i=-20.0°\text{C}$

Final temperature of water vapor needed, $T_f=120.0°\text{C}$

Formula used:

Heat energy transferred due to the difference in temperature is given by,

  $Q=mc\left(T_f-T_i\right)$

Where m is the mass of the substance, c is the specific heat, $T_f$ and $T_i$ are the final and initial temperatures, respectively.

The heat of fusion is given by

  $H_f=\frac{Q}{m}$

And, the heat of valorization is given by,

  $H_v=\frac{Q}{m}$

Where $Q$ is the quantity of heat, m is the mass of the substance

Calculation:

Now, the ice at $-20.0°\text{C}$ first will reduce its temperature to $0°\text{C}$ , then will convert from ice to water, then the temperature of water will rise to $100°\text{C}$ , then will again convert from liquid water to water vapor, and then the temperature of water vapor will raise to $120.0°\text{C}$ . Thus, the amount of thermal energy added is,

  $Q=m{c}_{ice}\left(0-\left(-20°\text{C}\right)\right)+m{H}_f+m{c}_{water}\left(100°\text{C}-0°\text{C}\right)+m{H}_v+m{c}_{steam}\left(120°\text{C}-100°\text{C}\right)$

It is known that,

  $\begin{array}{l}{c}_{ice}=2060\text{J/kg}\cdot \text{K}\\ H_f=3.34\times {10}^5\text{}\text{J/kg}\\ c_{water}=4180\text{J/kg}\cdot \text{K}\\ H_v=2.26\times {10}^6\text{}\text{J/kg}\\ c_{steam}=2020\text{J/kg}\cdot \text{K}\end{array}$

Using known and given values in above,

  $\begin{array}{c}Q=\left(0.01\right)\left[\left(\left(2060\right)\left(20\right)\right)+\left(3.34\times {10}^5\right)+\left(\left(4180\right)\left(100\right)\right)+\left(2.26\times {10}^6\right)+\left(\left(2020\right)\left(20\right)\right)\right]\\ =\left(0.01\right)\left[41200+334000+418000+2260000+40400\right]\\ =3.094\times {10}^4\text{J}\end{array}$

Conclusion:

Thus, the amount of thermal energy added is $3.094\times {10}^4\text{J}$ .