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How is the Hvap used to calculate the mass of liquid boiled by one kj of energy?
Wiki User
∙ 6y ago
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
Wiki User
∙ 6y ago
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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 of a solid melts?
Grams solid × mol/g × Hfusion
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 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 the Hvap used to calculate the mass of liquid boiled by 1 kJ of energy?
1kJ x 1/deltaHvap x g/mol liquid.
How is stoichiometry used to calculate energy absorbed when a mass liquid boils?
Grams liquid × mol/g × Hvap
How is Stoichiometry use to calculate energy absorbed when a mass of liquid boils?
Grams liquid × mol/g × Hvap
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 is stoichiometry used to calculate energy absorbed when a mass of a solid melts?
Grams solid × mol/g × Hfusion
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.
What do you call the amount of heat needed to transform mass of liquid to vapor?
It's called the melting point.
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
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.
How much energy is needed to evaporate 1g of water at room temperature?
1 Calorie per degree C. If room temp is 72 degrees C then 28 calories to get to boiling then the it takes 540 calories to water from liquid to gas, called the heat of vaporization. So add 540 to 28 and you get 568 calories. That's the energy to convert 1g of water at room temperature to 1g of steam at 100deg C. And if your room temperature is really 72 deg C you are in trouble - more probably degrees Fahrenheit. Many times, the units of energy are given in joules (J) and 1 calorie is equivalent to 4.184 joules. The specific heat of water is 4.184 J/g/deg, so to raise the temp of 1 g water from 25ºC to 100ºC will take (4.184 J/g/deg)(75 deg)(1g) = 314 J. Then to convert that 1 g of 100ºC water into steam, you need to use the ∆Hvap which is 2260 J/g. Thus, 2260 J/g x 1 g = 2260 J to evaporate the water. Total energy required is 314 J + 2260 J = 2574 J.
