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sineoko [7]
3 years ago
6

water has an unusually high heat of vaporization (-40+kj/mol), allowing it to be used for evaporative cooling. what do you think

would happen if a different compound was used for evaporative cooling - for example, methane (-8kj/mol)
Biology
1 answer:
Nataly_w [17]3 years ago
6 0

Answer:

A lesser amount of heat will be drawn up from the mass to be cooled per unit mole of the new liquid vaporized, compared to the amount of heat that will be drawn by vaporizing a unit mole of water. For the case of methane, five mole of methane needs to be vaporized to get the same cooling effect of a mole of water.

Explanation:

For a liquid like water to vaporize, it must draw a certain amount of energy from the surrounding in order for some of its molecules to gain enough energy to break out of the surface of the liquid as gas. If the heat of vaporization of water is about  -40 kJ/mol, it means that for a mole of water to change from liquid to gas, it must draw up 40 kJ worth of heat from its surrounding. The effect is that the surrounding mass losses 40 kJ worth of heat, and cools down. If we should now use a new liquid, say methane with heat of vaporization of about -8kJ/mol,  five moles of methane needs to be vaporized to get the same cooling effect of vaporizing a mole of water; since the heat that will be drawn by a mole of the methane is just bout 8 kJ.

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Explanation:

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'“Negative control” is a treatment that by definition is expected not to have any effect (neither positive effect, nor negative effect). “Positive control” is treatment with a well-known chemical that is known to produce the expected effect with the assay that you are studying. Application of an antagonist is not a negative control in your case. “Negative control” is condition that should be treated with the same solutions or buffers as your “treatment” condition, with the only difference that instead of the chemical that you investigate you should add just the solvent that was used to dissolve you chemical in the respective final concentration that you have in the “experimental treatment” condition. For example if your chemical is dissolved in DMSO – than the correct negative control will be to add to the medium/buffer just DMSO in the same final concentration that you reach with your “treatment” condition. One of the reasons of using such negative control is to verify that the solvent is having no effect in your assay. Note that among all treatment conditions (“negative control”, “positive control”, “experimental treatment you are investigating”) the volumes and the composition of the treatments that you are doing should be uniform: always treat with the same volume of medium or buffer, always containing the same concentration of the used solvent (e.g., DMSO). The only difference should be the presence or absence of the defined compound-treatments (agonist, antagonist, the chemical for the experimental investigation etc.).'

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Atanasov, Atanas. (2013). Re: Positive control and negative control. Retrieved from: https://www.researchgate.net/post/Positive_control_and_negative_control/515968f2d039b1fe50000025/citation/download.

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