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Question

An air parcel having initial temperature of 15°C rises adiabatically to 2 km height above the surface. What is the final temperature of the parcel?

The correct answer is
~-4.6°C

Adiabatic Ascent: Final Temperature Calculation

An air parcel rising adiabatically cools due to expansion. This calculation uses the Dry Adiabatic Lapse Rate (DALR), approximately $9.8 ^\circ C$ per kilometer, representing the temperature decrease with altitude.

Calculation Steps

  1. Convert Initial Temperature: Convert the initial surface temperature from Celsius to Kelvin for calculation.

    Given Initial Temperature ($T_1$): $15^\circ C$.

    Using the conversion formula $T(\text{K}) = T(^\circ C) + 273.15$:

    $ T_1 = 15 + 273.15 = 288.15 \text{ K} $

  2. Calculate Temperature Decrease: Determine the total temperature drop using the DALR and the altitude gained.

    Altitude Gain ($\Delta z$): $2$ km.

    Dry Adiabatic Lapse Rate (DALR): $9.8 ^\circ C/\text{km}$.

    Temperature Decrease ($\Delta T$) = DALR $\times \Delta z$:

    $ \Delta T = 9.8 ^\circ C/\text{km} \times 2 \text{ km} = 19.6 ^\circ C $

  3. Calculate Final Temperature (Kelvin): Subtract the calculated temperature decrease from the initial temperature in Kelvin.

    Final Temperature ($T_2$) in Kelvin:

    $ T_2 (\text{K}) = T_1 (\text{K}) - \Delta T $

    $ T_2 = 288.15 \text{ K} - 19.6 ^\circ C = 268.55 \text{ K} $

  4. Convert Final Temperature (Celsius): Convert the final temperature back to Celsius to match the options.

    Using the conversion formula $T(^\circ C) = T(\text{K}) - 273.15$:

    $ T_2 (^\circ C) = 268.55 \text{ K} - 273.15 $

    $ T_2 = -4.6 ^\circ C $

The final temperature of the air parcel is approximately $ -4.6 ^\circ C $. This corresponds to Option C.

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