amount of heat to evaporate water
We want to find Q so, 1  the specific heat of water is 4.2 J/kgC^-1. 2  i assume that the mass of the water is 1 kg. 3  the temperature difference in order to evaporate water is about 80C (100C - 20C assume that room temperature is 20C)  we substitute all in the 1  equation, Q = (4.2*10^3 J/kgC^-1) (1 kg) ( 2 80C) Q = 336000  J. Water evaporating takes quite a lot of heat away -- 540 calories per gram -- when it evaporates. That’s enough to cool down 540 grams of water by a degree, or 50 grams of water a little more than ten degrees. If you are not very careful to replace the lost heat energy during the evaporation, the temperature will go down. Follow-Up #3: ideal evaporation rate. Your answer to calculating the evaporation rate of water is of great interest to us, but you cite Zemansky and Dittman, Heat and Thermodynamics as your source. But I have that book (7th edition) and cannot find the equation or derivation. Use the formular δQ=cmΔt. which means the energy requirment of evaporate water equals its mass multiplies change in temperature and the specific heat capacity of water(which equals 4.2*10^3 J/(kg*℃)) For example: You want to boil 1 liter of water from 20℃ to 100℃ at STP. First , we calculate the mass of the water: That is, water has a high heat of vaporization, the amount of energy needed to change one gram of a liquid substance to a gas at constant temperature. Water’s heat of vaporization is around 540 cal/g at 100 °C, water's boiling point. Note that some molecules of water – ones that happen to have high kinetic energy – will escape from the surface of the water even at lower temperatures. In other words, water has a high specific heat capacity, which is defined as the amount of heat needed to raise the temperature of one gram of a substance by one degree Celsius. The amount of heat needed to raise the temperature of 1 g water by 1 °C is has its own name, the calorie. The latent heat of vaporization of liquid water is 2.5 MJ/kg. Just as an interesting comparison, the specific heat of liquid water is 4190 J/kg, so raising the temperature of water from the melting point to the boiling point takes 4190*100 = 419,000 J/kg or 0.419 MJ/kg, which is about 1/6 the energy it takes to boil it. 5 people found this useful. Example - Calculate heat required to evaporate 10 kg of water. The latent heat of evaporation for water is 2257 kJ/kg. The heat required to evaporate 10 kg can be calculated as. q = (2257 kJ/kg) (10 kg) = 22570 kJ. This three-part activity consists of an activity that student groups develop themselves, a given procedure, and an optional demonstration. First, students discuss examples of evaporation and then design and conduct their own test to find out whether heating water has an effect on the rate of evaporation. While waiting for their results, students conduct another evaporation activity using single drops of water on 2 paper towels, one of which is heated. Students should mention some way of adding heat to make water evaporate faster. Have students get into groups to design an experiment to test whether heat increases the rate of evaporation. Distribute the student activity sheet.
Around 540 cal/g at 100 °C.