# Amount of water vaporized when cooled down naturally

Suppose we have $$m_i$$ kg of liquid water at $$100 °C$$. This water cools down to $$20 °C$$. We want to estimate roughly the amount of water that would have to evaporate, in adiabatic conditions with the surrounding, so as to reach $$20 °C$$.

From first principle we have:

$$\int_{ini}^{final} d(mh)=-\int_0^t \dot{m_{out}}h_{out}dt$$

Assuming $$X$$ is the amount of water that evaporated we can approximate roughly that:

$$m_fh_f-m_ih_i=-Xh_{out}$$

Where $$h_{out}$$ would be the average enthalpy of water between initial state and final state. More, $$m_f = m_i - X$$, thus $$(m_i - X)h_f - m_ih_i =-Xh_{out}$$

So, $$X=m_i \left(\frac{h_f-h_i}{h_f-h_{out}}\right)$$

Now, I don't know what I am doing wrong, but as $$h_{out}$$ is an average value, then $$X$$ must be greater than $$m_i$$ which is not possible. Anyone can tell me what I am doing wrong?

• Could you please define your variables and describe how you're obtaining your equations. (For example, are you performing an energy balance?) You have sensible heat transfer going from 100°C to 20°C, but I'm not seeing where you incorporate the heat capacity/specific heat, which is the coefficient that links energy to a temperature change. Jun 28 at 23:03
• @Chemomechanics $d(mU)/dt = In - Out + Q + W$, nothing goes in, no heat transfer then $Q = 0$, no work done so $W = 0$ and because we have a liquid we can approximate $U$ as $H$. $X$ and $m$ are in $kg$. i stands for initial and f for final. Jun 28 at 23:17
• There are 3 forms of heat transfer, and those 3 forms all occur at the same time. Can you define the problem to the point where responders know what to do with each form of heat transfer? Jun 29 at 2:09

The simplest version of this problem is to assume that you have a rigid insulated container, initially with 1 kg liquid water at 100 C in the lower portion of the container and vacuum in the head space. The volume of the container is such that, in the final equilibrium state, the liquid water at 20 C plus the water vapor at 20 C exactly fills the container. Application of the 1st law of thermodynamics to this system gives: $$\Delta U=(1-X)u_L(20)+Xu_V(20)-u_L(100)=0$$where X is the final mass of vapor in the container (kg), and $$u_L(T)$$ and $$u_V(T)$$ are the internal energies of liquid water and water vapor, respectively, at temperature T (C). From the steam tables, $$u_L(100)=418.9\ kJ/kg$$, $$u_L(20)=83.94\ kJ/kg$$, and $$u_V(20)=2403\ kJ/kg$$. Solving for X yields X = 0.144 kg. So 14.4 % of the water evaporates.
You mention evaporation, but your equations don't represent this; that is, they don't contain the latent heat of vaporization. (In the case of evaporation, the outgoing specific enthalpy $$h_\text{out}$$ wouldn't be the average for the liquid water; the enthalpy of a gas is much higher than that of the corresponding liquid.)
Instead, your equations describe what happens if you simply move some amount of liquid water out of the system. This doesn't affect the system temperature, so the specific enthalpy $$h$$ of the water remains unchanged; $$h_\text{f}=h_\text{i}=h_\text{out}$$.
The final equation $$X=m_\text{i} \left(\frac{h_\text{f}-h_\text{i}}{h_\text{f}-h_\text{out}}\right)$$ thus contains the indeterminate $$\frac{h_\text{f}-h_\text{i}}{h_\text{f}-h_\text{out}}=\frac{0}{0}$$, and $$X$$ can take any value (while being constrained in practice to $$0\le X\le m_\text{i}$$).