text: how can you find the heat of vaporization of water according to the experimental values given? scenario:  a beaker of 300ml of water at 64 degrees celsius is placed into a vacuum bell.  pressure is relieved and the substance is allowed to boil.  when the bell is removed there are 284ml of water left and the temperature is 44 degrees celsius.  problems: assuming that there was no heat lost or gained through a temperature difference of the remaining air inside the bell jar during the time of boiling and that the rate of cooling and boiling was constant,  1. explain the drastic change in temperature of the water. 2. calculate the heat of vaporization of the water in cal/gram according to the energy change of the water that was evaporated. i don't have a calculator handy, but i'll help where i can.  evaporation is endothermic, meaning that it takes in heat.  the amount of heat taken in when one mole of water evaporates is its molar enthalpy of vaporization.  so, first, find out how much of the water evaporated.  assuming standard conditions, 1 ml of water weighs 1g.  so that means that 16g of water evaporated. (with me so far?)  now use that to find out that you have a bit less than on mole of water vapor.  where did the energy that the evaporating water use come from?  the other molecules of water.  you know that the amount of energy released when 284g of water decreases temperature is related in the formula q=mc(tfinal-tinitial).  (only the other 284g of water decreased temp...the other 16g gained temp.)  now you can figure out how much energy the water vapor absorbed when it changed into a gas.  take the energy # and divide by the # of moles... and you are set.   oh side note: since you are dropping the pressure, you do not need to worry about heating the water up inorder to make it boil.