Weather & ClimateRising Air and Stability

Saturated Air Cools More Slowly

Once a parcel is saturated, condensation releases latent heat, so it cools more slowly, about 6 °C per kilometer on average.

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Above the cloud base, a rising parcel is saturated, and every bit more cooling condenses some vapor. Condensing releases latent heat inside the parcel. That changes how fast it cools.

Key idea

A saturated parcel cools more slowly than a dry one: about 6 °C per km on average (the moist adiabatic lapse rate). It is about 4 °C per km in warm, humid air near the ground and approaches the dry rate (9.8 °C per km) in very cold air high up. Problems here use 6 °C per km unless they say otherwise.

Why it works

Expansion still cools the parcel by 9.8 °C for each km. But as it cools, vapor condenses, and each gram of vapor that condenses gives about 2500 J back to the parcel's air. That pays back part of the cooling. Warm saturated air condenses a lot of vapor for each degree of cooling (saturation climbs so steeply with temperature), so it gets a big payback and cools only about 4 °C per km. Cold air has little vapor left to condense, so it cools almost as fast as dry air.
Below the cloud base (1 km here), the parcel cools at 10 °C per km; above it, at 6 °C per km. The bend in the red line is where condensation begins.

Worked example: Climbing inside a cloud

A parcel is saturated at its cloud base, 1 km up, at 16 °C. It keeps rising to 5 km. What is its temperature? (Use 6 °C per km.)
  1. It climbs $5 - 1 = 4$ km while saturated.
  2. Cooling: $4 \times 6 = 24$ °C.
  3. $16 - 24 = -8$ °C. A dry parcel would have cooled 40 °C, to −24 °C.

Answer: −8 °C

Worked example: From the ground to the top of a cloud

Ground air at 26 °C with a dew point of 18 °C is lifted to 4 km. Use 10 °C per km (dry), 2 °C per km for the dew point and 6 °C per km (moist). Find its temperature at 4 km.
  1. Cloud base: the spread of 8 °C closes at 8 °C per km, so the base is at 1 km.
  2. At the base: $26 - 10 \times 1 = 16$ °C (and the dew point is $18 - 2 = 16$ °C).
  3. From 1 km to 4 km, moist: $16 - 6 \times 3 = -2$ °C.

Answer: −2 °C

Common mistake

Thinking the released latent heat warms a rising cloud parcel. The parcel still cools as it rises; the latent heat only slows the cooling, from 9.8 to about 6 °C per km. Check: in any moist-ascent answer, the parcel's temperature at the top must be lower than at the cloud base.
The moist rate depends on temperature and pressure (computed from the standard formula).
Air at the cloudMoist lapse rate (about)
30 °C, near sea level3.6 °C per km
20 °C, near sea level4.3 °C per km
0 °C, near sea level6.5 °C per km
−20 °C, about 5.5 km up7.7 °C per km
−40 °C, about 9 km up9.0 °C per km

Practice

Pass with 3 right answers, the last 2 in a row. Each right answer earns 2 XP.

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Sources and further reading

  • NOAA National Weather Service, JetStream online school for weather: The Atmosphere: Adiabatic processes and stability
  • Wallace and Hobbs, Atmospheric Science: An Introductory Survey, 2nd ed., section 3.5 (saturated adiabatic lapse rate)
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