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Determining stability from temperature readings

Short answer

Environmental lapse rate is 8.4 °C/km, less than dry adiabatic 9.8, so stable. A parcel lifted dry-adiabatically reaches 27.55 °C, colder than surrounding 27.9 °C, confirming stability. If 270 m temperature were 26.7 °C, ELR becomes 13.2 °C/km, unstable.

The air temperature measured at 20 m above ground is 30.0 degrees C and at 270 m it is 27.9 degrees C. (a) Calculate the environmental lapse rate and state the stability condition. (b) A parcel of air at 20 m is lifted dry-adiabatically to 270 m. Find its temperature and show that your conclusion about stability holds. (c) What would the stability be if the temperature at 270 m was 26.7 degrees C?

Take the dry adiabatic lapse rate Γd=9.8 ∘\Gamma_d = 9.8\ ^\circC/km =0.98 ∘= 0.98\ ^\circC/100 m.

(a) Environmental lapse rate

Γ=T1−T2z2−z1=30.0−27.9270−20=2.1250 ∘C/m=0.84 ∘C/100 m=8.4 ∘C/km\Gamma = \frac{T_1 - T_2}{z_2 - z_1} = \frac{30.0 - 27.9}{270 - 20} = \frac{2.1}{250}\ ^\circ\text{C/m} = 0.84\ ^\circ\text{C}/100\ \text{m} = 8.4\ ^\circ\text{C/km}

Since Γ=8.4<Γd=9.8 ∘\Gamma = 8.4 < \Gamma_d = 9.8\ ^\circC/km, the atmosphere is subadiabatic (stable).

(b) Parcel temperature

Tp=30.0−9.8×0.25=30.0−2.45=27.55 ∘CT_p = 30.0 - 9.8 \times 0.25 = 30.0 - 2.45 = 27.55\ ^\circ\text{C}

The surroundings at 270 m are 27.9 ∘^\circC. The parcel (27.55 ∘^\circC) is colder and denser than the surrounding air, so it sinks back to its original level. The atmosphere resists vertical motion, so it is stable.

(c) New reading

Γ=30.0−26.7250×100=1.32 ∘C/100 m=13.2 ∘C/km>9.8\Gamma = \frac{30.0 - 26.7}{250} \times 100 = 1.32\ ^\circ\text{C}/100\ \text{m} = 13.2\ ^\circ\text{C/km} > 9.8

The parcel would arrive at 27.55 ∘^\circC, warmer than the surroundings (26.7 ∘^\circC), so it keeps rising: unstable (superadiabatic).

Answer: (a) ELR = 8.4 ∘^\circC/km, stable; (b) parcel 27.55 ∘^\circC < 27.9 ∘^\circC, stable; (c) ELR = 13.2 ∘^\circC/km, unstable.

✓ Answer checked by Claude Opus 5.5 (high effort)

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