Respuesta :
To determine the time it takes to completely vaporize the given amount of water, we first determine the total heat that is being absorbed from the process. To do this, we need information on the latent heat of vaporization of water. This heat is being absorbed by the process of phase change without any change in the temperature of the system. For water, it is equal to 40.8 kJ / mol.
Total heat = 40.8 kJ / mol ( 1.50 mol ) = 61.2 kJ of heat is to be absorbed
Given the constant rate of 19.0 J/s supply of energy to the system, we determine the time as follows:
Time = 61.2 kJ ( 1000 J / 1 kJ ) / 19.0 J/s = 3221.05 s
Total heat = 40.8 kJ / mol ( 1.50 mol ) = 61.2 kJ of heat is to be absorbed
Given the constant rate of 19.0 J/s supply of energy to the system, we determine the time as follows:
Time = 61.2 kJ ( 1000 J / 1 kJ ) / 19.0 J/s = 3221.05 s
[tex]\boxed{{\text{3213}}{\text{.15 s}}}[/tex] will be required to convert 1.50 mol of water completely into steam.
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.
Substitute 1.50 mol for n and 40.7 kJ/mol for [tex]\Delta {H_{{\text{vap}}}}[/tex] in equation (1).
[tex]\begin{aligned}{\text{q}}&= \left( {1.50{\text{ mol}}} \right)\left( {\frac{{40.7{\text{ kJ}}}}{{1{\text{ mol}}}}} \right)\\&= 61.05{\text{ kJ}}\\\end{aligned}[/tex]
The amount of energy is to be converted into J. The conversion factor for this is,
[tex]1{\text{ kJ}} = {\text{1}}{{\text{0}}^3}{\text{ J}}[/tex]
Therefore the amount of energy can be calculated as follows:
[tex]\begin{aligned}{\text{q}} &= \left( {61.05{\text{ kJ}}} \right)\left( {\frac{{{{10}^3}{\text{ J}}}}{{1{\text{ kJ}}}}} \right)\\&= {\text{61050 J}}\\\end{aligned}[/tex]
The time required for conversion of water into steam is calculated as follows:
[tex]\begin{aligned}{\text{Time required}}&= \left( {61050{\text{ J}}} \right)\left( {\frac{{1{\text{ s}}}}{{19{\text{ J}}}}} \right)\\&= 3213.15{\text{ s}}\\\end{aligned}[/tex]
Learn more:
- Calculate the enthalpy change using Hess’s Law: https://brainly.com/question/11293201
- Find the enthalpy of decomposition of 1 mole of MgO: https://brainly.com/question/2416245
Answer details:
Grade: Senior School
Subject: Chemistry
Chapter: Thermodynamics
Keywords: enthalpy of vaporization, q, n, 1.50 mol, water, steam, 3213.15 s, 61050 J, 61.05 kJ, 40.7 kJ/mol, liquid state, vapour state, substance, time, amount of energy.
