Respuesta :
Given the energy obtained by the human body from a candy bar = 915 kJ
Energy required to vaporize 1 g of water at it's boiling point is referred to as heat of vaporization of water.
The heat of vaporization of water = 2260 J/g
Mass of water that can be vaporized by 915 kJ of heat = [tex]915 kJ * \frac{1000 J}{1kJ} *\frac{1 g}{2260 J} =405 g[/tex]
Density of water = 1 g/mL
Converting mass to volume in L of 405 g water:
[tex]405 g * \frac{1 mL}{1 g} * \frac{1 L}{1000 mL} = 0.405 L[/tex]
Therefore, 0.405 L of water can be vaporized
[tex]\boxed{{\text{0}}{\text{.405 L}}}[/tex] of water could be vaporized if the energy of 915 kJ was used.
Further explanation:
Enthalpy of vaporization
It is the amount of energy that is required to convert a substance from its liquid state to a vapor or gaseous state. It is also known as the latent heat of vaporization or heat of evaporation. It is represented by [tex]\Delta {H_{{\text{vap}}}}[/tex].
The expression for the heat of vaporization is as follows:
[tex]{\text{q}} = {\text{n}}\Delta {H_{{\text{vap}}}}[/tex] …… (1)
Here,
q is the energy of the substance.
n is the number of moles of the substance.
[tex]\Delta {H_{{\text{vap}}}}[/tex] is the heat of vaporization.
Rearrange equation (1) for n.
[tex]{\text{n}} = \frac{{\text{q}}}{{\Delta {H_{{\text{vap}}}}}}[/tex] …… (2)
The amount of energy is 915 kJ.
The heat of vaporization of water is 40.7 kJ/mol.
Substitute these values in equation (2).
[tex]\begin{aligned}{\text{n}} &= \left( {915{\text{ kJ}}} \right)\left( {\frac{{1{\text{ mol}}}}{{40.7{\text{ kJ}}}}} \right)\\&= 22.48{\text{ mol}}\\\end{aligned}[/tex]
The formula to calculate the moles of a substance is as follows:
[tex]{\text{Moles of substance}} = \dfrac{{{\text{Given mass of substance}}}}{{{\text{Molar mass of substance}}}}[/tex] …… (3)
Rearrange equation (3) to calculate the mass of substance.
[tex]{\text{Mass of substance}} = \left( {{\text{Moles of substance}}} \right)\left( {{\text{Molar mass of substance}}} \right)[/tex] …… (4)
Substitute 22.48 mol for moles of substance and 18.01 g/mol for the molar mass of substance in equation (4) to calculate the mass of water.
[tex]\begin{aligned}{\text{Mass of water}} &= \left( {{\text{22}}{\text{.48 mol}}} \right)\left( {\frac{{{\text{18}}{\text{.01 g}}}}{{1{\text{ mol}}}}} \right)\\&= 404.86{\text{ g}}\\\end{aligned}[/tex]
The formula to calculate the density of water is as follows:
[tex]{\text{Density of water}} = \dfrac{{{\text{Mass of water}}}}{{{\text{Volume of water}}}}[/tex] …… (5)
Rearrange equation (5) to calculate the volume of water.
[tex]{\text{Volume of water}}=\dfrac{{{\text{Mass of water}}}}{{{\text{Density of water}}}}[/tex] …… (6)
Substitute 1 g/mL for the density of water and 404.86 g for the mass of water in equation (6).
[tex]\begin{aligned}{\text{Volume of water}}&= \left( {404.86{\text{ g}}} \right)\left( {\frac{{1{\text{ mL}}}}{{1{\text{ g}}}}} \right)\\&= 404.86{\text{ mL}}\\\end{aligned}[/tex]
The volume of water is to be converted into L. The conversion factor for this is,
[tex]1{\text{ mL}} = {\text{1}}{{\text{0}}^{ - 3}}{\text{ L}}[/tex]
So the volume of water can be calculated as follows:
[tex]\begin{aligned}{\text{Volume of water}}&= \left( {404.86{\text{ mL}}} \right)\left( {\frac{{{{10}^{ - 3}}{\text{ L}}}}{{1{\text{ mL}}}}} \right)\\&= 0.40486{\text{ L}}\\&\approx {\text{0}}{\text{.405 L}}\\\end{aligned}[/tex]
Learn more:
- What is the difference between heat and temperature: https://brainly.com/question/914750
- Determine the process by which water enters into the atmosphere? https://brainly.com/question/2037060
Answer details:
Grade: Senior School
Subject: Chemistry
Chapter: Thermodynamics
Keywords: density, enthalpy of vaporization, volume of water, 0.405 L, 404.86 g, 22.48 mol, heat of evaporation, q, n, mass of water.
