<p>The zero point of the enthalpy is based on gaseous air and  <em>liquid</em> water at 0°C, whereas the water content in the air is <em>water vapor</em>. So to get from the defined zero point to the actual state, you have to add the latent heat to get the water from liquid into vapor.</p>

CC BY-SA 4.0
2023-11-04T11:23:13.690Z
2023-11-04T11:23:13.690Z
Oscar Lanzi
<p>In a lecture book we read about latent heat of moist air.</p>
<p>I would have guessed, that the total specific enthalpy of moist air, comprised of <span class="math-container">$m_d$</span> kg of dry air and <span class="math-container">$m_v$</span> kg of water vapor is (we have just a mixture of two gases...)</p>
<p><span class="math-container">$$h_{total} = \frac{c_p^d \cdot T \cdot m_d+c_p^v \cdot T \cdot m_v }{m_d+m_v}$$</span></p>
<p>Defining absolute humidity as
<span class="math-container">$$q = \frac{m_v}{m_d+m_v}$$</span></p>
<p>this is written as:</p>
<p><span class="math-container">$$h_{total} = c_p^d\cdot (1-q)\cdot T + c_p^v\cdot q\cdot T $$</span></p>
<p>In the lecture, however, we learned</p>
<p><span class="math-container">$$h_{total} = c_p^d\cdot (1-q)\cdot T + L \cdot q $$</span></p>
<p>with L denoting the enthalpy of vaporization (aka latent heat). I know, what latent heat physically means, but anyway, I do not understand why it is included in h for the gas.</p>
<p>My questions:</p>
<p>Q1: The lecture book says literally, that the enthalpy of pure water vapor is</p>
<p><span class="math-container">$$H_v = m_v \cdot L$$</span></p>
<p>so, in contrast to my equation, the <span class="math-container">$c_p^v$</span> is not included at all. This would mean, water vapor just contributes a constant to h which doesn't depend anymore on T - for me this sounds weird, since we are still talking about a gas...</p>
<p>Q2: Why is it required at all, to include L into the enthalpy of moist air? After liquid water has vaporized (for whatever reason) it is just a &quot;regular&quot; gas with another specific gas constant and also the mixture of those two gases behaves like an ideal gas to a good approximation. So, to make my question  more clear: If L for water is included in h, why isn't the enthalpy of vaporization of liquid dry air also included? Physically there is no difference, but it would be somehow ridiculous to include latent heat of liquid air into the enthalpy of the atmospheric gas.</p>
<p>I see no reason to do so - although I believe, that good arguments exist.</p>
<p>This is the copy from the book: Unfortunately, since it is a german book, D and L denote here vapor (&quot;Dampf&quot;) and dry air (&quot;Luft&quot;), respectively.
<a href="https://i.sstatic.net/FvsVa.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/FvsVa.png" alt="enter image description here" /></a></p>

<meteorology><atmosphere><atmosphere-modelling><thermodynamics><air>
Why is latent heat of water vapor included in specific enthalpy of moist air?
MichaelW
# Why is latent heat of water vapor included in specific enthalpy of moist air?

(Asked by: username_0 on 2023-04-15)

In a lecture book we read about latent heat of moist air.

I would have guessed, that the total specific enthalpy of moist air, comprised of $m\_d$ kg of dry air and $m\_v$ kg of water vapor is (we have just a mixture of two gases...)

$$h\_{total} = \frac{c\_p^d \cdot T \cdot m\_d+c\_p^v \cdot T \cdot m\_v }{m\_d+m\_v}$$

Defining absolute humidity as
$$q = \frac{m\_v}{m\_d+m\_v}$$

this is written as:

$$h\_{total} = c\_p^d\cdot (1-q)\cdot T + c\_p^v\cdot q\cdot T $$

In the lecture, however, we learned

$$h\_{total} = c\_p^d\cdot (1-q)\cdot T + L \cdot q $$

with L denoting the enthalpy of vaporization (aka latent heat). I know, what latent heat physically means, but anyway, I do not understand why it is included in h for the gas.

My questions:

Q1: The lecture book says literally, that the enthalpy of pure water vapor is

$$H\_v = m\_v \cdot L$$

so, in contrast to my equation, the $c\_p^v$ is not included at all. This would mean, water vapor just contributes a constant to h which doesn't depend anymore on T - for me this sounds weird, since we are still talking about a gas...

Q2: Why is it required at all, to include L into the enthalpy of moist air? After liquid water has vaporized (for whatever reason) it is just a "regular" gas with another specific gas constant and also the mixture of those two gases behaves like an ideal gas to a good approximation. So, to make my question more clear: If L for water is included in h, why isn't the enthalpy of vaporization of liquid dry air also included? Physically there is no difference, but it would be somehow ridiculous to include latent heat of liquid air into the enthalpy of the atmospheric gas.

I see no reason to do so - although I believe, that good arguments exist.

This is the copy from the book: Unfortunately, since it is a german book, D and L denote here vapor ("Dampf") and dry air ("Luft"), respectively.
[![enter image description here](https://i.sstatic.net/FvsVa.png)](https://i.sstatic.net/FvsVa.png)

**Tags:** meteorology, atmosphere, atmosphere-modelling, thermodynamics, air


---

## 1 answers

### Answer by username_1 on 2023-11-04. Score: 2

The zero point of the enthalpy is based on gaseous air and *liquid* water at 0°C, whereas the water content in the air is *water vapor*. So to get from the defined zero point to the actual state, you have to add the latent heat to get the water from liquid into vapor.


---

