Chapter 13, Problem 55AE

Interpretation Introduction

Interpretation:

Energy liberated for condensation of steam at $100°\text{C}$ to liquid at $30°\text{C}$ has to be calculated.

Concept Introduction:

Heat required by $\text{1 g}$ of a substance to increase the temperature by $1°\text{C}$ is termed as specific heat capacity of that substance. Energy absorbed or released by the substance is calculated as follows:

  $\text{Energy}=\left(\text{mass}\right)\left(\text{specific heat}\right)\left(\text{change in temperature}\right)$

Heat of condensation is the quantity of heat released on condensation of a steam to liquid. Every substance has different heat of condensation. It is represented as $-{\text{ΔH}}_{\text{v}}$ and calculated as follows:

  $\text{Energy}=-\left(\text{mass}\right)\left({\text{ΔH}}_{\text{v}}\right)$

Explanation of Solution

The formula to calculate mass of water is as follows:

  ${\text{moles of H}}_{\text{2}}\text{O}=\frac{{\text{mass of H}}_{\text{2}}\text{O}}{{\text{molar mass of H}}_{\text{2}}\text{O}}$        (1)

Rearrange equation (1) for mass of ${\text{H}}_{\text{2}}\text{O}$.

  ${\text{mass of H}}_{\text{2}}\text{O}=\left({\text{moles of H}}_{\text{2}}\text{O}\right)\left({\text{molar mass of H}}_{\text{2}}\text{O}\right)$        (2)

Substitute $50\text{ mol}$ for moles of ${\text{H}}_{\text{2}}\text{O}$ and $18.015\text{ g/mol}$ for molar mass of ${\text{H}}_{\text{2}}\text{O}$ in equation (2).

  $\begin{array}{c}{\text{mass of H}}_{\text{2}}\text{O}=\left(50\text{ mol}\right)\left(18.015\text{ g/mol}\right)\\ =900.75\text{ g}\end{array}$

The formula to calculate energy of condensation of steam to water is as follows:

  $\text{E}=\text{m}\cdot \text{c}\cdot \left({\text{T}}_{\text{2}}-{\text{T}}_{\text{1}}\right)$        (1)

Here,

$\text{E}$ is energy of water.

$\text{m}$ is mass of water.

$\text{c}$ is specific heat of water.

${\text{T}}_{\text{2}}-{\text{T}}_{\text{1}}$ is change in temperature.

Substitute $900.75\text{ g}$ for $\text{m}$, $4.184\text{ J/g }°\text{C}$ for $\text{c}$, $30°\text{C}$ for ${\text{T}}_{\text{2}}$ and $100°\text{C}$ for ${\text{T}}_1$ in equation (1).

  $\begin{array}{c}\text{E}=\left(900.75\text{ g}\right)\left(4.184\text{ J/g }°\text{C}\right)\left(30°\text{C}-100°\text{C}\right)\\ =\left(900.75\text{ g}\right)\left(\frac{4.184\text{ J}}{1\text{ g }°\text{C}}\right)\left(-70°\text{C}\right)\\ =-263811.66\text{ J}\end{array}$

The formula to calculate energy releasedfrom steam to form liquid is as follows:

  $\text{q}=-\text{m}\cdot {\text{ΔH}}_{\text{v}}$        (2)

Here,

$\text{q}$ is the energy released by water to condense steam

$\text{m}$ is the mass of steam

${\text{ΔH}}_{\text{v}}$ is the heat of vaporization

Substitute $900.75\text{ g}$ for $\text{m}$ and $2260\text{ J/g}$ for ${\text{ΔH}}_{\text{v}}$ in equation (2).

  $\begin{array}{c}\text{q}=-\left(900.75\text{ g}\right)\left(\frac{2260\text{ J}}{1\text{ g}}\right)\\ =-2035695\text{ J}\end{array}$

Total energy released for condensation of steam to water is as follows:

  $\text{Total energy}=\text{E}+\text{q}$        (3)

Substitute $-263811.66\text{ J}$ for $\text{E}$ and $-2035695\text{ J}$ for $\text{q}$ in equation (3).

  $\begin{array}{c}\text{Total energy}=-263811.66\text{ J}--2035695\text{ J}\\ =-2299506.66\text{ J}\end{array}$

Thus, energy liberated by steam to form liquid is $-2299506.66\text{ J}$.