Chapter 10, Problem 37QP

Calculate the heat absorbed when 542 g of ice at $-15.0°\text{C}$ melts and then the water is converted to steam $145°\text{C}\text{.}$ The specific heat of ice $2.03\text{J(g°C),}$ the heat of fusion of ice is $6.01\times {10}^3\text{J/mol,}$ the specific heat of water is $4.18\text{J(g°C),}$ the heat of vaporization of water is $4.07\times {10}^4\text{J/mol,}$  and the specific heat of steam is $2.02\text{J(g°C)}\text{.}$

Interpretation Introduction

Interpretation:

The heat absorbed when ice is converted into steam is to be determined.

Concept Introduction:

Phase change is the process by which ice is converted into steam. Each phase change occurs at a constant temperature along with a change in energy. This change in energy is called the heat of that process, $q$ .

The heat absorbed or released by a substance is calculated by the following equation:

$q=m\times C\times \Delta T$ ……(1)

Here, $m$ is the mass, $\Delta T$ is the temperature change, and $C$ is the specific heat capacity.

The expression to calculate the value of $\Delta T$ :

$\Delta T=T_2-T_1$ …..(2)

Here $T_1$ is the initial temperature and $T_2$ is the final temperature.

Substitute (2) in (1)

$q=m\times C\times \left(T_2-T_1\right)$ ……(3)

Answer to Problem 37QP

Solution:

The heat absorbed when ice is converted into steam is $\text{1697}\text{.26 kJ}$ .

Explanation of Solution

Given Information: Mass of ice is $542\text{ g}$ , the temperature of ice is $-15°\text{C}$ , specific heat of ice is $2.03{\text{ J g}}^{-1}°{\text{C}}^{-1}$ , the heat of fusion of ice is $6.01\times {10}^3{\text{ J mol}}^{-1}$ , the specific heat of water is $\text{4}{\text{.18 J g}}^{-1}°{\text{C}}^{-1}$ , the heat of vaporization of water is $4.07\times {10}^4{\text{ J mol}}^{-1}$ and the specific heat of steam is $2.02{\text{ J g}}^{-1}°{\text{C}}^{-1}$ .

Using equation (3) to calculate the heat required to raise the temperature of ice from $-15°\text{C}$ to $0.0°\text{C}$ .

Substitute $542\text{ g}$ for $m$ , $2.03{\text{ J g}}^{-1}°{\text{C}}^{-1}$ for $C$ as the specific heat of ice, $0.0°\text{C}$ for $T_2$ and $-15°\text{C}$ for $T_1$ in equation (3):

$\begin{array}{c}{q}_1=542\text{ g}\times 2.03{\text{ J g}}^{-1}°{\text{C}}^{-1}\times \left(0.0°\text{C}-\left(-{15}^{°}\text{C }\right)\right)\\ \text{=16503}\text{.9 J}\end{array}$

Calculate the heat change for the melting of ice by multiplying the molar heat of fusion by the number of moles, $n$ .

$\begin{array}{c}{q}_2=n\times \text{molar heat of fusion}\\ \text{=}\left(\frac{\text{542 g}}{\text{18}{\text{.02 g mol}}^{-1}}\right)\times \left(6.01\times {10}^3{\text{ J mol}}^{-1}\right)\\ =180766.92\text{ J}\end{array}$

Using equation (3) to calculate the heat required to raise the temperature of water from $0.0°\text{C}$ to $100°\text{C}$ .

$q_3=m\times C\times \Delta T$

Substitute $542\text{ g}$ for $m$ , $\text{4}\text{.18 J/g}°\text{C}$ for $C$ as the specific heat of ice, $100°\text{C}$ for $T_2$ and $0.0°\text{C}$ for $T_1$ in equation (3):

$\begin{array}{c}{q}_3=542\text{ g}\times 4.18{\text{ J g}}^{-1}°{\text{C}}^{-1}\times \left(100.0°\text{C}-0.0°\text{C}\right)\\ =226\text{556 J}\end{array}$

Now, calculate the heat change for the evaporation of water by multiplying the molar heat of vaporization by the number of moles.

$\begin{array}{c}{q}_4=n\times \text{molar heat of vaporization}\\ \text{=}\left(\frac{\text{542 g}}{\text{18}{\text{.02 g mol}}^{-1}}\right)\times \left(4.07\times {10}^4{\text{ J mol}}^{-1}\right)\\ =1224162.04\text{ J}\end{array}$

Using equation (3) to calculate the heat required to raise the temperature of water from $100°\text{C}$ to $145°\text{C}$ .

Substitute $542\text{ g}$ for $m$ , $\text{2}{\text{.02 J g}}^{-1}°{\text{C}}^{-1}$ for $C$ as the specific heat of ice, $145°\text{C}$ for $T_2$ and $100°\text{C}$ for $T_1$ in equation (3):

$\begin{array}{c}{q}_5=542\text{ g}\times 2.02{\text{ J g}}^{-1}\text{C}{}^{-1}\times \left(145°\text{C}-100°\text{C}\right)\\ =49\text{267}\text{.8 J}\end{array}$

Calculate the total heat absorbed in the entire process by adding the heats for the individual steps as follows.

$\begin{array}{c}{q}_{total}=q_1+q_2+q_3+q_4+q_5\\ =16503.9\text{ J }+180766.92\text{ J }+226556\text{ J }+1224162.04\text{ J }+49267.8\text{ J}\\ =1697256.66\text{ J}\\ =1697.26\text{ kJ}\end{array}$

So, the total heat absorbed in the process is $\text{1697}\text{.26 kJ}$ .

Conclusion

The heat absorbed when ice is converted into steam is $\text{1697}\text{.26 kJ}$ .