Question

At 1 atm, how much energy is required to heat 75.0 g of H2O(s) at –22.0 °C to H2O(g) at 145.0 °C? Helpful constants can be found here.

Quantity	per gram	per mole
Enthalpy of fusion	333.6 J/g	6010. J/mol
Enthalpy of vaporization	2257 J/g	40660 J/mol
Specific heat of solid H2O (ice)	2.087 J/(g·°C) *	37.60 J/(mol·°C) *
Specific heat of liquid H2O (water)	4.184 J/(g·°C) *	75.37 J/(mol·°C) *
Specific heat of gaseous H2O (steam)	2.000 J/(g·°C) *	36.03 J/(mol·°C) *

Answer 1

Concepts and reason

Enthalpy of the reaction is equal to the amount of energy absorbed in a reaction.

Calculate the amount of heat required by using the following formula.

$q = m \times c \times \Delta T$

Here, q is heat released by the solution that equal to $n \times \Delta {\rm{H}}$ . Here, n indicates that number of moles. M is mass of the sample, $c$ is the specific heat, and $\Delta T$ is temperature change.

Fundamentals

The amount of heat required to raise one degree of temperature of one unit mass of sample is called as specific heat.

Calculate the energy in the form of heat as shown below.

$\begin{array}{c}\\{q_1} = m \times {C_{{\rm{water}}\left( g \right)}} \times \Delta T\\\\ = 75.0g \times 2.087{\rm{ J/g}}{{\rm{.}}^{\rm{o}}}{\rm{C }} \times {\rm{ }}\left[ {0{{\rm{ }}^{\rm{o}}}{\rm{C}} - \left( { - {{22.0}^{\rm{o}}}{\rm{C}}} \right)} \right]\\\\ = 3443{\rm{ J = 3}}{\rm{.443 kJ}}\\\end{array}$

The enthalpy of the change is as follows:

$\begin{array}{c}\\{q_2} = m \times \Delta {H_{{\rm{fusion}}}}\\\\ = {\rm{75}}{\rm{.0 g}} \times 333.6{\rm{ J/g}}\\\\ = 25020{\rm{ J = 25}}{\rm{.020 kJ}}\\\end{array}$

The enthalpy of the change is as follows:

$\begin{array}{c}\\{q_3} = m \times {C_{{\rm{water}}\left( g \right)}} \times \Delta T\\\\ = 75.0g \times 4.184{\rm{ J/g}}{{\rm{.}}^{\rm{o}}}{\rm{C }} \times {\rm{ }}\left[ {100{{\rm{ }}^{\rm{o}}}{\rm{C}} - \left( {{0^{\rm{o}}}{\rm{C}}} \right)} \right]\\\\ = 31380{\rm{ J = 31}}{\rm{.380 kJ}}\\\end{array}$

The enthalpy of the vaporization is as follows:

$\begin{array}{c}\\{q_4} = m \times \Delta {H_{{\rm{vapor}}}}\\\\ = {\rm{75}}{\rm{.0 g}} \times 2257{\rm{ J/g}}\\\\ = 169275{\rm{ J = 169}}{\rm{.275 kJ}}\\\end{array}$

The enthalpy of the change is as follows:

$\begin{array}{c}\\{q_5} = m \times {C_{{\rm{water}}\left( g \right)}} \times \Delta T\\\\ = 75.0g \times 2.000{\rm{ J/g}}{{\rm{.}}^{\rm{o}}}{\rm{C }} \times {\rm{ }}\left[ {145{{\rm{ }}^{\rm{o}}}{\rm{C}} - \left( {{{100}^{\rm{o}}}{\rm{C}}} \right)} \right]\\\\ = 6750{\rm{ J = 6}}{\rm{.750 kJ}}\\\end{array}$

The enthalpy of the change is as follows:

$\begin{array}{c}\\q = {q_1} + {q_2} + {q_3} + {q_4} + {q_5}\\\\ = 3.443{\rm{ kJ + 25}}{\rm{.020 kJ + 31}}{\rm{.380 kJ + 169}}{\rm{.275 kJ + 6}}{\rm{.750 kJ}}\\\\ = 235.9{\rm{ kJ }}\\\\{\rm{ = 235}}{\rm{.9 kJ}}\\\end{array}$

Ans:

Energy required to heat the 75.0 g of water is ${\rm{235}}{\rm{.9 kJ}}$ .