# Heat lost in evaporation at constant temperature

I ran an experiment evaporating salt water at different temperatures and concentrations. I’m trying to figure out how much energy is needed to maintain water at a specific temperature even when its evaporating. I have tried a calculating the heat loss by evaporation but since the mass is decreasing during evaporation, both mass and heat would change, but temperature would have to remain constant. I am going crazy trying to figure this out. Using the equation here would give me watts https://www.google.ca/amp/s/www.engineeringtoolbox.com/amp/evaporation-water-surface-d_690.html but it doesnt account for change in mass or temperature of the water.

Not a complete answer, but this should help clarify things for you:

You wrote "since the mass is decreasing during evaporation, both mass and heat would change". But note that the change in mass per se does not affect the needed heat flow. For instance, consider pure water. Here, the change in mass would not affect the rate at which heat is lost due to evaporation (assuming other conditions, like surface area, temperature, the partial pressure of water vapor in the air above the water, wind speed, etc., don't change), and thus would not affect the heat flow needed to maintain the temperature. This is because you're not trying to change the temperature of the water (in which case its mass would be relevant), you're just trying to maintain an energy balance of heat flow in and heat flow out.

Hence, regardless of the volume of water, for a specific evaporation rate, the heat flow needed to maintain constant temperature would be the same, since the latter is determined only by the rate at which heat is lost due to evaporation.

However, since you are working with salt water solutions, their concentration would increase with time, which would decrease the evaporation rate and thus decrease the heat flow needed to maintain the temperature.

So you have two choices:

(1) Just calculate the heat flow as a function of concentration (i.e., for each solution, determine the initial heat flow needed). If the volume of water is large relative to the surface area, treating the concentration as a constant could be a good approximation for short periods of time.

(2) Use the evaporation rate as a function of concentration to determine the concentration as a function of time. I.e., derive a time-dependent expression. This is trickier.

I'm not sure from your question which you want to do, since your concern seemed to be the change in the total mass of the water, not the concentration. Note also that the equation you found in the linked article doesn't include a variable for concentration.

• thank you but wouldn’t mass matter though because I need temperature to stay constant and the same ammount of heat with a smaller mass would have a higher temperature. Also I’m wondering if maintaining higher temperatures would require more energy. I’m going to set aside the concentration aspect of this experiment for now. – Jim Apr 28 '19 at 3:25
• Assuming the only loss of thermal energy (E_t) is due to evaporation, the mass wouldn't matter, since what determines the heat flow needed to maintain the temp. is the rate at which E_t is lost. I.e, if you are losing 1 cal/min due to evaporation, then you only need to add 1 cal/min to maintain the same temp., independent of the mass. OTOH, in genl., you would need more heat flow to maintain temp. with a larger mass b/c (assuming a const. surface area:vol ratio), a larger mass would lose E_t more quickly. But in your problem you only specified loss of E_t due to evaporation. – theorist Apr 28 '19 at 4:10
• Furthermore, if the temperature of the water is the same as the ambient temperature, then having a larger mass (and thus a larger total surface area, assuming a constant surface area: volume ratio) wouldn't matter when it comes to maintaining temperature, i.e., the only E_t loss would be due to evaporation, since the temperature of the air and the container would be equal to the water temperature. – theorist Apr 28 '19 at 5:07