# Calculating final temperature of humid air and cool stone

I'm trying to gauge the efficiency of a night-cooled stone to condense water from the humid air at seashores at noon. I'm assuming the stone is thermally insulated once it's in thermal equilibrium with the lowest possible temperature at night. Then at noon, the humid air is passed through the stone and the (assume thermally isolated) system now consists of hot, humid air and the cool stone.

I understand that at lower temperatures, air holds lesser vapor, so if hot humid air is condensed, some vapor will condense.

[EDIT] I already know the mass of water I am considering since I use the maximum vapor in a meter cube of saturated air possible at the temperatures. The problem I am having is which quantities to include on the left side of the equation for the vapor condensing. I'm using the specific heat capacity of vapor, air and stone to try and come up with the final temperature, so that I can see how much vapor will air condense at that temperature by subtracting the after-cooling holding capacity per meter-cubed from the initial one.

The problem here that I must use the latent heat of vaporisation for the energy lost to the stone for the water condensed, but then the vapor is at a temperature lower than 100 Celsius so should I use the specific heat capacity of vapor or latent heat or specific heat capacity of water?! I just got confused as hell.

As $$q_{vapor}+q_{air}=-q_{stone}$$, I came up with this $$m_{vapor}c_{Pvapor}(T_{1}-T_{F})+m_{air}c_{Pair}(T_{1}-T_{F})=-m_{stone}c_{Pstone}(T_{1}-T_{F})$$

Something just seems wrong to me here because of the vapor temperature being below boiling point of water. I know that temperature is AVERAGE internal energy of a mass of substance and all. However it still confuses me. Someone clear it up for me.

• You're approaching this from the wrong direction. Your first step should be to estimate the maximum amount of water you can get. What is the initial relative humidity of the air at the starting temperature, and what is the equilibrium vapor pressure of water at this temperature? This, along with the atmospheric pressure will allow you to calculate the amount of water vapor in the room air initially. Commented Apr 22, 2021 at 12:13
• Check out the edit Commented Apr 22, 2021 at 13:57
• So you have a fixed mass of air, a fixed mass of water, and a fixed mass of stone? What are these values? Also, what is the initial temperature of the air and the stone? Commented Apr 22, 2021 at 14:44

Let $$m_a$$, $$m_w$$, and $$m_s$$ represent the masses of (bone dry) air, the mass of water, and the mass of the stone, respectively. Let $$T_0$$ be the initial temperature of the air and $$T_s$$ be the initial temperature of the stone. Then, assuming all the water condenses, the change in internal energy unit mass of the bone dry air is $$m_aC_a(T_f-T_0)$$ the change in internal energy per unit mass of the water is $$m_w(-\lambda_0+C_{Lw}(T_f-T_0))$$, and the change in internal energy of the stone is $$m_sC_s(T_f-T_s)$$where $$\lambda_0$$ is the change in internal energy per unit mass of water at $$T_0$$ in going from water vapor to liquid water, $$C_{Lw}$$ is the specific heat capacity of liquid water. So the internal energy balance for this system should read: $$m_aC_a(T_f-T_0)+m_w(-\lambda_0+C_{Lw}(T_f-T_0))+m_sC_s(T_f-T_s)=0$$Again, this assumes that all the water condenses.
• If you have "steam tables," they have tabulated values for that. Alternately, you can look up the heat of vaporization of water as a function of temperature on Google, take the value at that temperature, and subtract $\Delta (PV)$; the latter is roughly equal to (RT_0/M), where R is the universal gas constant and M is the molar mass of water. Commented Apr 25, 2021 at 18:01