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A sample of H2O with a mass of 46.0 grams has a temperature of 100.0C How many joules of energy are necessary to boil the water?
Wiki User
∙ 11y ago
According to Einsteins theory, the fact that he had only one teste clearly adentifies that the degrees that you have provided are deniable therefore the answer should would and will be always zero degres divided by the percentage of the aqua which is h20
Wiki User
∙ 11y ago
Wiki User
∙ 9y agoTo boil liquid water that is already at 100º C requires an extra 550 calories per gram, called the latent heat of vaporisation.
So 46 gm would require 25,300 calories, and at 4.2 Joules per calorie that is 106,260 Joules. That might be done by 1000 watts running for 106 seconds.
Wiki User
∙ 11y agoIf the sample is at 100 0C, at standard pressure, no supplementary heating is necessary.
Wiki User
∙ 7y agoUsing 2260 J/g as the ∆Hvaporization
q = 46.0 g x 2260 J/g = 103,960 J = 104 kJ (to 3 sig. figs)
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The average Kinetic energy of the atoms in the sample will increase as the sample is heated.
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The specific heat capacity of water is 4186 joules per kilogram. That is to raise 1kg or 1 litre of water by 1 degree you will need to add 4186 joules of energy. So for 15grams over 25 degrees you will need 4186/1000*15*15 joules.
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How many joules of energy are necessary to heat a sample of water with a mass of 46 g?
46 calories (or 192, 464 joules) for each Celsius degree.
How many joules of energy are necessary to heat a sample of water with a mass of 46.0 grams for 0.0?
46 calories (or 192, 464 joules) for each Celsius degree.
A sample of H½O with a mass of 46.0 grams has a temperature of?
96.3 Joules
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46 calories (or 192, 464 joules) for each Celsius degree.
A sample of h2o with a mass of 46.0 grams has a temperature of 100 c how many joules of energy are necessary to boil the water?
Using 2260 J/g as the ∆Hvaporizationq = 46.0 g x 2260 J/g = 103,960 J = 104 kJ (to 3 sig. figs)
How many joules of energy are necessary to heat a sample of water with a mass of 46.0 grams from 0.0 celsius to 100.0?
419.1 Joules are required to heat one gram of liquid water from 0.01 degC to 100 deg C. So the answer is 419.1*46 = 19278.6
1. 10 J of heat are removed from a gas sample while it is being compressed by a piston that does 20 J of work. What is the change in thermal energy of the gas Does the temperature of the gas increase?
10 JoulesConservation of energy, assuming there are no other losses in the system, and 20 Joules are introduced by compression, and 10 Joules are removed by heat transfer, the remaining 10 Joules must be absorbed as increased thermal energy of the gas.
A sample of H20 with a mass of 46.0 grams has a temperature of 100 c How many joules of energy are necessary to boil the water?
The heat of vaporization is about 2258 joules/gram46 x 2258 = 103,868 J ~ 103.9 kJ(Using the rounded value 2260 joules per gram, about 104 kJ)it is already at boiling point so in theory zero energy is required.I think the question is asking what energy is required to free the water from its liquid state to the vapour state.At 101325Pa, (1 atmosphere in old money), the latent heat of evaporation of water is 2256.7 J/g.So 2256.7 * 46 = 103808 Joules
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24 Joules
What happened to the kinetic energy of the particles in a sample of gas as the temperature of the sample increase?
The average Kinetic energy of the atoms in the sample will increase as the sample is heated.
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When a sample of a substance absorbs thermal energy, its temperature rises.
