Question

1. a hot - air balloon including the envelope, gondola, burner fuel and one passenger has a total mass of 450 kg. the air outside the balloon is at 20 °c and has a density of 1.29 kg/m³. the air inside the envelope is heated to a temperature of 120 °c, at which it has a density of 0.90 kg/m³. what volume must the envelope expand in order to lift the balloon into the air?

Explanation:

Step1: Recall Archimedes' Principle

The buoyant force \( F_b \) equals the weight of the displaced air, and for the balloon to lift, the buoyant force must equal the total weight of the balloon (mass \( m = 450 \, \text{kg} \)) plus the weight of the heated air inside the envelope. Let \( V \) be the volume of the envelope (and displaced air), \(
ho_{\text{outside}} = 1.29 \, \text{kg/m}^3 \) (density of outside air), \(
ho_{\text{inside}} = 0.90 \, \text{kg/m}^3 \) (density of inside air), \( g \) be acceleration due to gravity.

Buoyant force: \( F_b=
ho_{\text{outside}} V g \)
Total weight to lift: \( W=(m +
ho_{\text{inside}} V)g \)

For equilibrium (lift), \( F_b = W \):
\(
ho_{\text{outside}} V g=(m +
ho_{\text{inside}} V)g \)

Step2: Cancel \( g \) and solve for \( V \)

Cancel \( g \) from both sides:
\(
ho_{\text{outside}} V=m +
ho_{\text{inside}} V \)

Rearrange terms to isolate \( V \):
\(
ho_{\text{outside}} V -
ho_{\text{inside}} V=m \)
\( V(
ho_{\text{outside}} -
ho_{\text{inside}})=m \)

Solve for \( V \):
\( V = \frac{m}{
ho_{\text{outside}} -
ho_{\text{inside}}} \)

Step3: Substitute values

Substitute \( m = 450 \, \text{kg} \), \(
ho_{\text{outside}} = 1.29 \, \text{kg/m}^3 \), \(
ho_{\text{inside}} = 0.90 \, \text{kg/m}^3 \):
\( V=\frac{450}{1.29 - 0.90} \)
Calculate denominator: \( 1.29 - 0.90 = 0.39 \)
Then \( V=\frac{450}{0.39}\approx1153.85 \, \text{m}^3 \)

Answer:

The volume of the envelope must be approximately \(\boldsymbol{1154 \, \text{m}^3}\) (or \(\frac{450}{0.39} \approx 1153.85 \, \text{m}^3\)).