# Water cooling by evaporation

For many years of drinking water has been cooled in hot climates by evaporating it from the surfaces of canvas bags or porous clay pots. How many grams of water can be cooled from $$\pu{35 ^\circ C}$$ to $$\pu{20 ^\circ C}$$ by the evaporation of $$\pu{60 g}$$ of water?

Heat of vaporisation of water in this temperature range is $$\pu{2.4 kJ g-1}$$. Specific heat of water is $$\pu{4.18 J g-1 K-1}$$.

I'm pretty confuse by what the question is asking for. This is my attempt for the evaporation of $$\pu{60 g}$$ of water:

$$q = \pu{60 g} \cdot \pu{2.4 kJ g-1} = \pu{144 J}$$

I then subbed this energy unit ($$\pu{144 J}$$) to the heat change formula to find the mass of water. Why am I wrong?

• What answer did you finally get and why do you think it's wrong? – Ivan Neretin Dec 8 '17 at 14:40
• @IvanNeretin I think it's wrong because 144J is the energy needed overcome the forces to form the bond at 100 degrees C (evaporation), but I have no idea on how to find how many grams can be formed from 35 to 20 degrees C – user175089 Dec 8 '17 at 15:15
• You said something about some heat change formula, so in fact you did have an idea. What made you abandon it? – Ivan Neretin Dec 8 '17 at 15:30
• @IvanNeretin I abandon it because after looking through the question again and again, I cannot substitute 144KJ into the heat temperature change formula as it make no sense. With that, this means I need to find the amount of heat energy (q) to sub it in to find the mass which is what I am unclear about . – user175089 Dec 8 '17 at 15:41
• Two questions that might bring you closer to solve this yourself: First of all, check the units. Is it J or kJ? Second, how much energy does water need to be heated from 20 to 35 °C? – mhchem Dec 8 '17 at 16:12

It helps to write down the formulas, before you plug in the numbers.

Heat you need to extract to cool the water:

$$Q=m_1 c_p\Delta T$$

Heat needed to evaporate water:

$$Q=m_2L_v$$

So:

$$m_1 c_p\Delta T=m_2L_v$$

So:

$$m_1=\frac{m_2L_v}{c_p \Delta T}$$

Now plug in the numbers.

I'm pretty confuse by what the question is asking for.

Because you are confused, I'll try to explain without any formulas. Please follow this step for step. Do not continue with the next step if you did not understand the previous one.

• When water evaporates, it needs energy for that.
• The task assumes that water evaporating from a clay pot will cool the water in clay pot. Part of the energy of the water in the pot (part of its 35 °C) will be used for evaporation of the 60 g.
• The task simplifies things enormously. We have no indication that we would take care of the pot itself, of the surrounding air or anything.
• How much energy does the evaporation of 60 g of water need? You already calculated that. But you need to double check your units! Is it J or kJ? It is 144 kJ.
• Now, this energy is taken from the water inside the pot. The water has some energy (because it is 35 °C). Can we calculate that? We could, it we knew how much water it was, but we don't. But you could write down a formula that takes the mass of the water and results in the energy of that water. What would this be?
• We can do the same thing with water of 20 °C.
• We could then create the formula for "How much more energy does 35 °C water have in comparison to 20 °C water.
• Okay so far? But we are lucky the energy difference between water of 35 °C and 20 °C is the same as between water of 15 °C and 0 °C. There is a constant that sais "How much energy is needed to heat one gram of water by 1 °C". You are lucky, you already have it mentioned in the original question.
• So, if we need 4.18 J to heat one gram of water by 1 °C, how much energy do we need to heat one gram of water by 15 °C? Do you have that number? It is 15 times 4.18 J = 62.7 J.
• The energy needed to heat water by 15 °C is the same as we need to remove if the want to cool water by 15 °C.
• Okay, we know how much energy is needed to evaporate the 60 g. That is 144 kJ.
• We know how much energy we need to remove from the inner water to cool one gram of it by 15 °C. That is 62.7 J. If there were 2 grams in the pot, it would need double that, 125.4 J. If there were 3 grams in the pot, it would need 188.1 kJ to cool it down by 15 °C.
• How many grams of water could we cool by 15 °C with the removal of 144 kJ? That is how often does 62.7 J fit into 144 kJ? Be careful with the units! How often does 62.7 J fit into 144,000 J?

If you understood every step of this "manual", you should try to answer the same question with formulas. This will be much faster. But, of course, you need to understand the underlying concept first. This is why I wrote this step by step instructions.