How much energy must be removed to cool a volume of air 1 km high by 20 km wide x 30 km deep by 5°C? (The density of air is about 1.29 g/L; it requires 1.02 J to heat up 1 g of air 1K.) (C = 75.291 J/mol.K for H₂O; q = C_cal$\Delta$T)

Question

Quizplus · Accepted Answer

First, calculate the volume of air: $1 \, \text{km} \times 20 \, \text{km} \times 30 \, \text{km} = 600 \, \text{km}^3 = 600 \times 10^9 \, \text{m}^3$. Since 1 L = $10^{-3} \, \text{m}^3$, the volume in liters is $600 \times 10^9 \times 10^3 = 600 \times 10^{12} \, \text{L}$. Given the density of air is 1.29 g/L, the mass of the air is $600 \times 10^{12} \, \text{L} \times 1.29 \, \text{g/L} = 774 \times 10^{12} \, \text{g}$. To cool this air by 5°C (or 5K, since the change in temperature is the same in both units), and given it requires 1.02 J to heat 1 g of air by 1K, the energy required is $774 \times 10^{12} \, \text{g} \times 5 \, \text{K} \times 1.02 \, \text{J/gK} = 3.96 \times 10^{15} \, \text{J}$, or $4.0 \times 10^{12} \, \text{kJ}$, since $1 \, \text{kJ} = 10^3 \, \text{J}$.

How Much Energy Must Be Removed to Cool a Volume Δ\DeltaΔ

How Much Energy Must Be Removed to Cool a Volume $\Delta$