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Eddi Din [679]
3 years ago
6

How long would it take for 1.50 mol of water at 100.0 ∘c to be converted completely into steam if heat were added at a constant

rate of 19.0 j/s ?
Chemistry
2 answers:
blagie [28]3 years ago
6 0

\boxed{{\text{3213}}{\text{.15 s}}} 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 \Delta {H_{{\text{vap}}}}.

The expression for the heat of vaporization is as follows:

{\text{q}} = {\text{n}}\Delta {H_{{\text{vap}}}}                                                                …… (1)

Here,

q is the energy of the substance.

n is the number of moles of the substance.

\Delta {H_{{\text{vap}}}} is the heat of vaporization.

Substitute 1.50 mol for n and 40.7 kJ/mol for \Delta {H_{{\text{vap}}}} in equation (1).

 \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}

The amount of energy is to be converted into J. The conversion factor for this is,

 1{\text{ kJ}} = {\text{1}}{{\text{0}}^3}{\text{ J}}

Therefore the amount of energy can be calculated as follows:

\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}

The time required for conversion of water into steam is calculated as follows:

\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}  

Learn more:

  1. Calculate the enthalpy change using Hess’s Law: brainly.com/question/11293201
  2. Find the enthalpy of decomposition of 1 mole of MgO: 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.

Aleksandr-060686 [28]3 years ago
5 0
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
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