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?
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.
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} $
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 $
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} $
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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