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What is the best definition of the enthalpy of vaporization delta Hvap?
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How is the delta Hvap used to calculate the mass of liquid boiled by 1 kj of energy?
1kJ x 1/Hvap x g/mol liquid
How is stoichiometry used to calculate energy absorbed when a mass liquid boils?
Grams liquid × mol/g × Hvap
How is the Hvap used to calculate the mass of liquid boiled by 1 kJ of energy?
1kJ x 1/deltaHvap x g/mol liquid.
How does the vapor pressure of water at 10 C compare with its vapor pressure at 50 C?
The vapor pressure of water at 50ºC will be greater than that at 10ºC because of the added energy and thus greater movement of the water molecules. If one knows the ∆Hvap at a given temperature, one can calculate the vapor pressure at another temperature. This uses the Clausius-Clapeyron (sp?) equation. It turns out the vapor pressure of water at 10º is 9.2 mm Hg, and that at 50º is 92.5 mm Hg.
How much heat is needed to vaporize 7.24 ml of sweat from your skin at 25 degree c?
If you assume that the density of sweat is the same as that of water (1g/ml) and that the specific heat of sweat is the same as that of water (4.184 J/g/deg), then one can estimate the heat required to evaporate 7.24 ml of sweat at 25 degrees Cheat to raise 7.24 ml from 25º to 100º = q = mC∆T = (7.24g)(4.184 J/g/deg)(75deg) = 2272 J heat to evaporate 7.24 ml = q = m∆Hvap = (7.24g)(2260 J/g) = 16,362 J Total heat = 2272 J + 16,362 J = 18,634 J = 18.6 kJ
How is the delta Hvap used to calculate the mass of liquid boiled by 1 kj of energy?
1kJ x 1/Hvap x g/mol liquid
How is the delta Hvap used to calculate the mass liquid boiled by 1 kj of energy?
1kJ x 1/Hvap x g/mol liquid
How is stoichiometry used to calculate energy absorbed when a mass liquid boils?
Grams liquid × mol/g × Hvap
How much energy in kilojoules is required to convert 155 mL of water at its boiling point from liquid to vapor delta H vap H20 equals plus 40.7 kilojoules per mole?
The density of water @ 100oC (boiling point) is about 0.958 g/ml. First we need to convert the 155 ml to mass by multiply by the density.155 ml * (0.958 g/ml) = 148.49 gramsNext convert the grams to moles by dividing by the molecular weight of water, which is 18 g/mol:148.49 grams /(18 g/mol) = 8.25 mol of H2OFinally multiply the moles of water by the heat of vaporization (Hvap) to get the final answer:8.25 mol * (40.7 kJ/mol) = 335.775 kJ
How is Stoichiometry use to calculate energy absorbed when a mass of liquid boils?
Grams liquid × mol/g × Hvap
What is the calculated atmospheric pressure in mm Hg if the boiling point of water boils at 81 degrees Celsius?
This would be a fairly simple experiment to do. You place your water in a chamber which is pressurized (or de-pressurized) to the desired degree of pressure, then slowly heat it with a Bunsen burner until it starts to boil; a thermomenter in the water will then tell you the temperature. If you just want the information, and don't want to do the experiment yourself, information about the boiling and freezing point of water at all different temperatures and pressures is given in what is known as a phase diagram. This can be found by way of Google under "water phase diagram" or in the Handbook of Physics and Chemistry.
What do you call the amount of heat needed to transform mass of liquid to vapor?
It's called the melting point.
How is the Hvap used to calculate the mass of liquid boiled by one kj of energy?
The Hvap is usually given as J/g or J/mole for example. Let's say you have it in J/g. For water this is about 2260 J/g, meaning it takes 2260 joules to vaporize 1 g of water. Set up an equation using dimensional analysis: 1 kJ x 1000 J/kg x 1 g/2260 J = 0.44 g
How is stoichiometry used to calculate energy absorbed when a mass of a solid melts?
Grams solid × mol/g × Hfusion
How is the Hvap used to calculate the mass of liquid boiled by 1 kJ of energy?
1kJ x 1/deltaHvap x g/mol liquid.
How much energy is removed when 10.0g of water is cooled from steam at 133.0 C to liquid at 53.0 C?
The ∆Hvap for water is 2260 J/g. Thus, 10 g x 2260 J/g = 22600 Joules or 22.6 kJoules
How many btus to raise 1 pound of water from room temp deg to steam 213 deg?
First convert 1 lb of water to lb-moles which is 0.055 lb-moles (you'll need this later). This problem can be broken into 3 steps:(1) You need to detemine how much heat is needed to raise room temperature water (68oF) to 212oF. This can be used using the heat capacity of water which at room temperature is 1 Btu/lboF. So the amount of heat needed for this is:Q1 = m*Cp*ΔT= (1 lb)*(1 Btu/lboF)*(212 - 68oF)= 144 Btu(2) Next you need to account for the phase change. The water changes to steam at 212oF. You use the heat of vaporization which you can look up in any Chemistry or Chemical Engineering Handbook. The Hvap that I found is 17493.5 Btu/lb-mole.Q2 = n(lb-moles)*Hvap= (0.055 lb-moles)*(17493.5 Btu/lb-mole)= 972.64 Btu(3) Next you need to find out how much heat is needed to raise the temperature of the steam from 212 to 213oF. You can look up the heat capacity of steam at 212oF to be 0.485 Btu/lboF.Q3 = m*Cp*ΔT= (1 lb)*(0.485 Btu/lboF)*(213-212oF)= 0.485 BtuTo find the total heat needed add Q1+Q2+Q3 (144+972.64+0.485) =1117.12 Btu
