Chapter 12.2, Problem 23PP

To determine

The amount of thermal energy to be added to a substance for a given change in temperature.

Answer to Problem 23PP

The amount of thermal energy to be added is $9.402\times {10}^5\text{ J}$

Explanation of Solution

Given:

The temperature of $3.00\times {10}^2 \text{g}$ of ice is changed from $-30°\text{C}$ to water vapour at $130°\text{C}$ by adding a specific amount of thermal energy.

Formula Used:

When temperature of an object is changed from $T_A$ (in Kelvin) to $T_B$ , there is always a gain or loss of thermal energy.

Here two change in states occur, which are melting of ice and changing of water into vapour. Hence the thermal energy consists of energy involved in melting of solid, energy for temperature changes and energy for converting water into vapor.

The energy for melting of ice of mass m is given by,

  $Q_{melt,ice}=m{H}_{\text{f}}$

In the above equation, the parameter $H_{\text{f}}$ represents the heat of fusion of ice.

The energy for changing water to vapor is given by,

  $Q_v=m{H}_{\text{v}}$

In the above equation, the parameter $H_v$ represents the heat of fusion of ice.

The energy for causing temperature change is dependent on the temperature change, mass of the object $m$ and specific heat $C$ , and is given by the expression,

  $Q_t=mC\left(T_B-T_A\right)$

Calculation:

Here, ice is having a mass of $m\text{=3}.00\times \text{1}{0}^{\text{2}}\text{g}$ . The temperature of ice is changed from

  $-30°\text{C}$ to water vapor at $130°\text{C}$ . The specific heat of ice is $C_i=2060\text{ J/kg}\text{.K}$ . The heat of fusion of ice is $H_{\text{f}}=3.34\times {10}^5\text{ J/kg}$ . The heat of vaporization is $H_{\text{v}}=2.26\times {10}^6\text{ J/kg}$ . The specific heat of water is $C_w=4180\text{ J/kg}\text{.K}$ .The specific heat of vapour is $C_s=2020\text{ J/kg}\text{.K}$

Firstly, the thermal energy for converting ice to water is to be calculated which is given by the sum of thermal energy for melting of ice and then temperature change from $T_1=-\text{3}0°\text{C}$ to $T_2=0°\text{C}$ as

  $\begin{array}{c}{Q}_A=m{C}_i\left(T_1-T_2\right)+m{H}_{\text{f}}\\ =0.300\times 2060\left(0-\left(-30\right)\right)+\left(0.300\right)\left(3.34\times {10}^5\right)\\ =1.187\times {10}^5\text{ J}\end{array}$

Now, the thermal energy for converting ice at $T_2=0°\text{C}$ to water at $T_3=100°\text{C}$ is given by,

  $\begin{array}{c}{Q}_B=m{C}_w\left(T_3-T_2\right)\\ =0.3\times 4180\times \left(100-0\right)\\ =1.254\times {10}^5\text{ J}\end{array}$

Thirdly, the thermal energy for converting water to vapor at $T_3=100°\text{C}$ to water at $T_4=130°\text{C}$ is the sum of thermal energy for converting water to vapor and the energy for causing the temperature change, which is calculated as,

  $\begin{array}{c}{Q}_C=m{C}_s\left(T_3-T_4\right)+m{H}_{\text{v}}\\ =0.300\times 2020\left(130-\left(100\right)\right)+\left(0.300\right)\left(2.26\times {10}^6\right)\\ =6.9618\times {10}^5\text{ J}\end{array}$

Therefore, the total energy for the given conversion is given by,

  $\begin{array}{c}Q=Q_A+Q_B+Q_C\\ =1.187\times {10}^5+1.254\times {10}^5\text{ +}6.9618\times {10}^5\\ =9.402\times {10}^5\text{}\text{J}\end{array}$

Conclusion:

Thus, the thermal energy required to raise the temperature of ice to vapor for the given range is $9.402\times {10}^5\text{}\text{J}$ .