Question

How much heat energy is required to heat a 100 g sample of liquid water from 30 °C to water vapor at 110 °C?The specific heat of liquid water is 4.184 J/(g K) and water vapor is 2.008 J/(g K). The heat of vaporization for water is 2259 J/g and the heat of fusion is 334.72 J/g

Answer 1

`Q_T=257.776kJ`

Step-by-step explanation:

The total heat energy required is expressed according to the formula:

`\begin{gathered} Q_T=Q_{\text{vap,w}}+Q_w \\ \end{gathered}`

where:

Qvap is the heat energy absorbed by the vapor

Qw is the heat energy absorbed by the water

Get the heat energy absorbed by the vapor at 100 degrees

`\begin{gathered} Q_{\text{vap,w}}=n_w*\triangle H_(vap,w) \\ Q_{\text{vap,w}}=\frac{mass}{molar\text{ mass}}*\triangle H_(vap,w) \\ \end{gathered}`

Given the following parameters;

Mass of water = 100g

Molar mass of water (H2O) = 18.015g/mol

Hvap,w = 2259 J/g = 40.8 kJ/mol

Substitute the given parameters into the formula to have:

`\begin{gathered} Q_{\text{vap,w}}=\frac{100\cancel{g}}{18.015\cancel{g}\cancel{\text{mol}^(-1)}^{}}*\frac{40.8kJ}{\cancel{\text{mol}}} \\ Q_{\text{vap,w}}=226.48kJ \end{gathered}`

Get the heat absorbed by the water from 30 to 100 degrees and from 100 to 110 degrees using the formula below. Note that the water vapor is being heated without any phase changes, so we will be utilizing the specific heat capacity of water vapor.

`\begin{gathered} Q_w=m_wc_w(\triangle\theta)_w+m_wc_w\triangle\theta \\ Q_w=(100*4.184*(100-30))+(100g*2.008(J)/(g^oC)*(110-100)^oC) \\ Q_w=(100*4.184*70)+(100*2.008*10) \\ Q_w=29,288+2008 \\ Q_w=31296\text{Joules} \\ Q_w=31.296kJ \\ \end{gathered}`

Get the total heat energy required;

`\begin{gathered} Q_T=226.48kJ+31.296kJ \\ Q_T=257.776kJ \end{gathered}`