SCIENCE · INVESTIGATE AND EXPLAIN

Compare a lifted parcel with its environment

Track idealized parcel cooling separately from the surrounding temperature profile.

Learning outcome: Calculate parcel and environmental temperatures at a shared height and distinguish a prescribed lift from a predicted trajectory.

Starting recap

A lapse rate is a temperature decrease per height. The environmental rate describes surroundings; a parcel cooling rule describes a hypothetical displaced air sample.

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Try

At the default 1 km height, the parcel is 20.2°C while the environment is 23°C, so the temperature difference is -2.8°C.

Open the experiment

Predict

Does changing the environmental lapse rate change this model’s prescribed parcel cooling curve?

Your prediction and reason

Experiment

Keep the surface temperature–dew-point difference at 8°C and compare environmental decreases of 7 and 8°C/km. Read parcel and environment at 1 and 3 km.

Lift a parcel through a temperature profile

Worked teaching model · synthetic values · Synthetic parcel and environment both start at 30 °C. Prescribed cooling: unsaturated parcel 9.8 °C/km, its dew point approximately 1.8 °C/km, saturated parcel a fixed teaching approximation of 6 °C/km. Actual moist cooling varies with pressure and temperature. Height spans 0–3 km above the starting level. No mixing, virtual-temperature correction, condensate loading, pressure solver, lift force, time evolution or precipitation. Temperature differences are only a buoyancy proxy, not CAPE or an actual forecast.

Surface temperature–dew-point difference: 8 °C Environmental temperature decrease: 7 °C/km

Reset keeps your prediction, captures and practice history.

Conditionally unstable lapse-rate comparison. This compares assumed lapse rates, not the weather outcome. The post-saturation parcel/environment temperature equality is 3.80 km (beyond this 3 km plot). The temperature equality is not a virtual-temperature-corrected level of free convection.

Compare parcel and environment at the same height
Height above start	Parcel	Environment	Parcel minus environment	Parcel stage
0.00 km	30.00 °C	30.00 °C	0.00 °C	Unsaturated approximation
0.50 km	25.10 °C	26.50 °C	-1.40 °C	Unsaturated approximation
1.00 km · condensation level	20.20 °C	23.00 °C	-2.80 °C	Saturated approximation
1.50 km	17.20 °C	19.50 °C	-2.30 °C	Saturated approximation
2.00 km	14.20 °C	16.00 °C	-1.80 °C	Saturated approximation
2.50 km	11.20 °C	12.50 °C	-1.30 °C	Saturated approximation
3.00 km	8.20 °C	9.00 °C	-0.80 °C	Saturated approximation

Positive temperature difference means warmer, negative means colder. At equal pressure and comparable composition this suggests a buoyancy tendency; actual density also depends on moisture and suspended condensate. This model supplies neither vertical speed nor the force needed to lift a colder parcel.

The initial gap is 8 °C. Its approximate closing rate is 9.8 − 1.8 = 8 °C/km, giving 1.00 km (1000 m). Rising air expands as surrounding pressure falls. Unlike the preceding humidity state comparison, pressure is not held fixed along this conceptual ascent. Cloud formation and precipitation require additional microphysics and observations.

Approximate lifting condensation level 1.00 km. At 3 km parcel minus environment = -0.80 °C.

Conditionally unstable lapse-rate comparison. Condensation level comes from the closing temperature–dew-point gap: 8/(9.8 − 1.8) km. Saturation does not guarantee that the parcel is warmer than its environment or that rain develops.

Observation to attach to a capture

Captures preserve recipes and replay their calculated results. They are synthetic model evidence, not physical measurements.

Understand

Two curves answer different questions

The environment is the surrounding atmosphere. A parcel is a hypothetical air sample displaced through it. A profile of environmental temperature is not automatically the temperature path that the parcel follows. This investigation starts both at 30°C at height zero, then assigns them separate rules.

The control called Environmental temperature decrease specifies b in degrees Celsius per kilometer. Its profile is T_env = 30 − bz, with height z in kilometers. At the default b = 7, the surroundings are 23°C at 1 km and 9°C at 3 km. This is a supplied straight-line profile, not a measured sounding.

Before saturation, the parcel follows T_parcel = 30 − 9.8z. An unsaturated parcel can contain water vapor; the term dry adiabatic does not mean it contains none. Adiabatic refers to the idealization of no heat exchange with the surroundings. Expansion during ascent can cool the parcel even without external heat loss. NOAA introduction to parcel theory.

Compare at the same height

With the default surface temperature–dew-point difference of 8°C, the model reaches saturation at 1 km. Parcel temperature there is 30 − 9.8 = 20.2°C. The environment is 30 − 7 = 23°C. Subtract environment from parcel: ΔT = 20.2 − 23 = −2.8°C.

The negative sign means the parcel is cooler in this comparison. It does not mean an air temperature below zero, and it is not a numerical acceleration. The displayed contrast is a temperature-only proxy for buoyancy that omits moisture-related density corrections and other effects.

On paper draw two vertical columns labeled parcel and environment. At each common height, write the two temperatures and connect them with a horizontal comparison line. This prevents accidentally comparing a surface parcel temperature with an environmental temperature at another level.

Follow the rule above saturation

Above the approximate saturation height, the parcel cools at a fixed 6°C/km in this teaching model. Starting from 20.2°C at 1 km, another 2 km gives 20.2 − 6 × 2 = 8.2°C at 3 km. Against the default 9°C environment, the contrast is −0.8°C.

The reduced cooling rate represents the influence of condensation and latent heat in a simplified way. A real moist adiabatic lapse rate varies with temperature and pressure; 6°C/km is not a universal constant. The environmental rate is a separate quantity again. NWS definitions of environmental and adiabatic lapse rates.

The model uses its prescribed rates rather than computing pressure, water content and latent heating from a complete thermodynamic solver. It conceptually compares parcel and environment at a common pressure at each level, but supplies no exact pressure profile.

Change the surroundings without changing the parcel rule

Keep the surface gap at 8°C and change b from 7 to 8°C/km. The parcel curve remains unchanged, while the environment becomes 22°C at 1 km and 6°C at 3 km. The contrasts become −1.8°C and +2.2°C. Capture both settings and identify which curve moved.

This controlled comparison shows why the initial surface temperature alone cannot describe stability through a layer. It also shows why a single phrase such as “warm air rises” cannot replace a shared-height comparison with explicitly stated assumptions.

The plot’s horizontal or vertical progression through altitude is not a clock. There is no lifting speed, force or mechanism that guarantees the parcel reaches the top. The calculation asks what the prescribed path would look like if the parcel were displaced through those levels.

Transfer to an unsaturated segment

For a paper-only check, consider a parcel that remains unsaturated through 0.5 km and an environment decreasing 5°C/km. Both start at 30°C. Find the parcel and environmental temperatures and their contrast.

Check the parcel-cooling transfer

The parcel is 30 − 9.8 × 0.5 = 25.1°C. The environment is 30 − 5 × 0.5 = 27.5°C, giving ΔT = −2.4°C. This is a cooler-parcel temperature comparison at a common height, not a calculated sinking speed or complete density assessment.

Apply

Public practice checks a specific claim. It does not establish independent mastery.

Distinguish the parcel from the environment, show one shared-height subtraction and explain why the curves do not predict a lifting speed.

Your explanation and limitations

Continue

Completion requires a saved prediction, at least one capture and a reflection. Presence checks do not grade your reasoning.

Skipping retains experiments and awards no mastery.

Previous: Interpret saturation limits without hiding the result Next: Estimate a lifting condensation level

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