Latent heat of vaporisation

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Heixi
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#1
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Why is the calculated value of Lv lower than the actual value ?
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Callicious
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Could you post the question and post what you've tried?
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Driving_Mad
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I assume this is a standard question.

Things to think about: is all the energy supplied (e.g. from the heater) transferred to the object being vaporised. Is there any of this energy being dissipated to the surroundings (e.g. due to a lack of insulation).
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Heixi
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(Original post by Driving_Mad)
I assume this is a standard question.

Things to think about: is all the energy supplied (e.g. from the heater) transferred to the object being vaporised. Is there any of this energy being dissipated to the surroundings (e.g. due to a lack of insulation).
I don’t think that would be it . If you use like a Liebig condenser or something the thermal energy losses would be eliminated
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Heixi
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(Original post by Callicious)
Could you post the question and post what you've tried?
Say you have an enclosed Liebig condenser and you heat something . You would measure the time taken to boil it , measure the mass and you would calculate specific latent heat by measuring voltage and current .
I don’t think heat or energy transfer would have an impact on the latent heat of vaporisation so what else could you say would be a factor ?
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Callicious
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(Original post by Heixi)
Say you have an enclosed Liebig condenser and you heat something . You would measure the time taken to boil it , measure the mass and you would calculate specific latent heat by measuring voltage and current .
I don’t think heat or energy transfer would have an impact on the latent heat of vaporisation so what else could you say would be a factor ?
The condenser method doesn't necessitate knowing the method by which the fluid used to produce the vapor is heated (https://aapt.scitation.org/doi/pdf/10.1119/1.1933207) it seems. What matters in that case is the amount of energy the vapour produces on condensation. You could heat it with the most inefficient awful process possible, but as long as the same mass of vapour reaches the condenser and room temperature stays the same, it's all irrelevant, it seems (according to that one source I found that references a method using that condenser specifically.)

I wouldn't expect the condenser to rely on the method by which the vapour was produced- just by how it's cooled, and that's unique to the condenser.

With regard to the condenser, though. To have your calculated value lower than the actual value, there'd have to be losses somewhere in your condenser apparatus (i.e. condensation isn't delivering the "correct" heat.)

That's easily reconcilable: radiative losses, conductive losses to the air, you name it- it's there.
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Heixi
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(Original post by Callicious)
The condenser method doesn't necessitate knowing the method by which the fluid used to produce the vapor is heated (https://aapt.scitation.org/doi/pdf/10.1119/1.1933207) it seems. What matters in that case is the amount of energy the vapour produces on condensation. You could heat it with the most inefficient awful process possible, but as long as the same mass of vapour reaches the condenser and room temperature stays the same, it's all irrelevant, it seems (according to that one source I found that references a method using that condenser specifically.)

I wouldn't expect the condenser to rely on the method by which the vapour was produced- just by how it's cooled, and that's unique to the condenser.

With regard to the condenser, though. To have your calculated value lower than the actual value, there'd have to be losses somewhere in your condenser apparatus (i.e. condensation isn't delivering the "correct" heat.)

That's easily reconcilable: radiative losses, conductive losses to the air, you name it- it's there.
I see , thanks
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