A high-altitude balloon is launched from an elevation of 350 m and rises until it bursts at an elevation of 10000 m. The pressure in the ballon is constant at 110 kPa, and the volume expands as the pressure outside the balloon decreases, from 0.2 m3 to 0.6 m3 . The balloon and its payload have a mass of 3 kg. The air inside the balloon loses 350 kJ as it rises. Determine the heat loss

Δ Q = 110 kJ

Explanation:

given,

height of elevation, h₁ = 350 m

height at which it burst,h₂ = 10000 m

Pressure in the balloon = 110 kPa

Volume of balloon= V₁ = 0.2 m³

                                 V₂ = 0.6 m³

Payload mass = 3 Kg

Loss of pressure = 350 kJ

Heat loss = ?

Change in Potential of the balloon

Δ P E = m g (h₂ - h₁)

Δ P E = 3 x 9.81 x (10000- 350)

         = 284 kJ

Air Loss in balloon = 350 kJ

Work done when the volume of balloon expand

W = pΔV

W = 110 x (0.6-0.2)

W = 44 kJ

heat loss calculation

From conservation of energy

Δ Q + Δ U = Δ x + W

Δ Q is the heat loss

Δ Q = Δ x + W - Δ U

Δ Q = 350 + 44 - 284

Δ Q = 110 kJ

∴ Heat lost is equal to 110 kJ

a.) travels faster in solids because the particles are closer together.

air is a gas water is a liquid