Question

How much energy is required to convert $100.0$ g of water at ${\text{20}}{\text{.0}}\,{\,^{\text{o}}}{\text{C}}$ completely to steam at ${\text{100}}{\text{.0}}\,{\,^{\text{o}}}{\text{C}}$?

Answer 1

The conversion of liquid water to steam takes place via change in temperature. At a time only one change occurs, change in temperature or change in phase. By using heat formula we can determine the heat change during temperature change and during phase change. By adding the heat required on both processes we can determine the total energy required.

Complete answer:
We have to convert the liquid water into steam. This conversion takes place by heating which increases the temperature. The process of conversion is shown as follows:
$\mathop {{\text{liq}}{\text{.water}}}\limits_{20.0{\,^{\text{o}}}{\text{C}}} \,\mathop \to \limits^1 \mathop {{\text{liq}}{\text{.water}}}\limits_{100.0{\,^{\text{o}}}{\text{C}}} \mathop \to \limits^2 \mathop {{\text{steam}}}\limits_{100.0{\,^{\text{o}}}{\text{C}}}$
In the first step, the liquid water boils so, the temperature of the liquid water increases from ${\text{20}}{\text{.0}}\,{\,^{\text{o}}}{\text{C}}$ to${\text{100}}{\text{.0}}\,{\,^{\text{o}}}{\text{C}}$.
The formula used to determine the heat change during heating of liquid water is as follows:
${\text{q}}\,{\text{ = }}\,{\text{m}}{{\text{C}}_{\text{p}}}{\Delta T}$
Where,
${\text{q}}$is the heat.
${\text{m}}$is the mass of the substance
${{\text{C}}_{\text{p}}}$is the heat capacity of the substance
${\Delta T}$is the change in temperature
Heat capacity of water at constant pressure, ${{\text{C}}_{\text{p}}}$ is ${\text{4}}{\text{.184}}\,{\text{J}}\,{\,^{\text{o}}}{{\text{C}}^{ - 1}}\,{{\text{g}}^{ - 1}}$. So, the heat required to increase the temperature of $100.0$ g of water is,
${{\text{q}}_1}\,{\text{ = }}\,{\text{m}}{{\text{C}}_{\text{p}}}{\Delta T}$
OR
${{\text{q}}_1}\,{\text{ = }}\,{\text{m}}{{\text{C}}_{\text{p}}}\left( {{{\text{T}}_{{\text{final}}}} - {{\text{T}}_{{\text{initial}}}}} \right)$
On substituting $100.0$g for the mass of water, ${\text{4}}{\text{.184}}\,{\text{J}}\,{\,^{\text{o}}}{{\text{C}}^{ - 1}}\,{{\text{g}}^{ - 1}}$for the heat capacity of water, ${\text{20}}{\text{.0}}\,{\,^{\text{o}}}{\text{C}}$ for initial temperature and ${\text{100}}{\text{.0}}\,{\,^{\text{o}}}{\text{C}}$ for final temperature.
${{\text{q}}_1}\,{\text{ = }}\,100.0\,{\text{g}} \times {\text{4}}{\text{.184}}\,{\text{J}}\,{\,^{\text{o}}}{{\text{C}}^{ - 1}}\,{{\text{g}}^{ - 1}}\left( {{\text{100}}{\text{.0}}\,{\,^{\text{o}}}{\text{C}} - {\text{20}}{\text{.0}}\,{\,^{\text{o}}}{\text{C}}} \right)$

${{\text{q}}_1}\,{\text{ = }}\,100.0\,{\text{g}} \times {\text{4}}{\text{.184}}\,{\text{J}}\,\, \times 8{\text{0}}{\text{.0}}$
${{\text{q}}_1}\,{\text{ = }}\,33472\,{\text{J}}$
So, the heat required to increase the temperature of water is $33472\,{\text{J}}$.
In the second step, the liquid water changes into steam and the temperature of the water does not change.
The formula used to determine the heat change during constant temperature phase transition is as follows:
${{\text{q}}_2}\,{\text{ = }}\,{m\Delta }{{\text{H}}_{{\text{vap}}}}$
Where,
${\Delta }{{\text{H}}_{{\text{vap}}}}$ is the enthalpy change during vaporization.
Enthalpy change of vaporization for water is ${\text{2260}}\,{\text{J}}\,\,{{\text{g}}^{ - 1}}$. So, the heat required to change the phase of $100.0$ g of water is,
${{\text{q}}_2}\,{\text{ = }}\,{m\Delta }{{\text{H}}_{{\text{vap}}}}$
On substituting $100.0$g for the mass of water, ${\text{2260}}\,{\text{J}}\,\,{{\text{g}}^{ - 1}}$for the enthalpy change of vaporization.
${{\text{q}}_2}\,{\text{ = }}\,100.0\,{\text{g}} \times {\text{2260}}\,{\text{J}}\,\,{{\text{g}}^{ - 1}}$
${{\text{q}}_2}\,{\text{ = }}\,{\text{226000}}\,{\text{J}}\,$
So, the heat required to change the phase of water is ${\text{226000}}\,{\text{J}}$.
The total heat required for the conversion of liquid water into steam is the sum of heat required in the first and second step.
${\text{q}}\,\,{\text{ = }}\,\,{{\text{q}}_1} + {{\text{q}}_2}$
${\text{q}}\,{\text{ = }}\,{\text{226000}}\,{\text{J}}\, + {\text{33472}}\,{\text{J}}\,$
${\text{q}}\,{\text{ = }}\,{\text{2}}59472\,{\text{J}}\,$

Therefore ${\text{2}}59472\,{\text{J}}$energy is required to convert $100.0$ g of water at ${\text{20}}{\text{.0}}\,{\,^{\text{o}}}{\text{C}}$ completely to steam at ${\text{100}}{\text{.0}}\,{\,^{\text{o}}}{\text{C}}$.

Note: The heat required to increase the temperature of a substance by one degree Celsius is known as heat capacity. The heat required to increase the temperature of one gram substance by one degree Celsius is known as specific heat capacity. The molar heat capacity of water at constant pressure, ${{\text{C}}_{\text{p}}}$ is ${\text{7}}{\text{.5}}\,{\text{J}}\,\,{{\text{K}}^{ - 1}}\,{\text{mo}}{{\text{l}}^{ - 1}}$ it means the energy required to increase the temperature of one mole of water by one kelvin is ${\text{7}}{\text{.5J}}$ . The unit of mass of compound moles are very important. When the heat is given the sign if q remains positive. When heat is released the sign of heat remains negative.