Question

When a parcel of dry air rises at a rate of 2 cm s$^{-1}$ vertically, what should be the rate of heating per unit mass in J s$^{-1}$ kg$^{-1}$ (due to radiation, conduction, etc.) in order to maintain the air parcel at a constant temperature? (Consider $g = 9.8 \, \text{m/s}^2$, rounded off to three decimal places).

Hint:

For constant temperature ascent, the radiative heating must exactly balance adiabatic cooling.

Answer 1

Correct Answer: 0.195

Step 1: Recall first law of thermodynamics for rising air.
The dry adiabatic lapse rate is: \[ \Gamma_d = \frac{g}{c_p} \] where $c_p \approx 1005 \, \text{J/kg/K}$.
Step 2: Calculate cooling rate due to adiabatic ascent.
Vertical velocity: \[ w = 2 \, \text{cm/s} = 0.02 \, \text{m/s} \] Cooling rate per unit time: \[ \frac{dT}{dt} = - \Gamma_d . w = - \frac{g}{c_p} . w \] \[ = - \frac{9.8}{1005} \times 0.02 = -0.000195 \, K/s \]
Step 3: Convert to heating rate per unit mass.
Energy required to offset cooling: \[ Q = c_p . \left|\frac{dT}{dt}\right| = 1005 \times 0.000195 = 0.196 \, J \, s^{-1} \, kg^{-1} \] Final Answer: \[ \boxed{0.196 \, J \, s^{-1} \, kg^{-1}} \]