Question

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)

Answer 1

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.