(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A hot air balloon has a volume of 1600m^{3}. The balloon, its passengers, basket and other equipment have a mass of 500kg. Using the formula below, find the average temperature the air in the balloon must be raised to for it to lift off the ground.

Density of surrounding air = 1.3kgm^{-3}.

2. Relevant equations

[tex]\rho=\rho_{0}\dfrac{273}{273+\Delta T}[/tex]

where [tex]\rho[/tex] = new density, [tex]\rho_{0}[/tex] = original density

3. The attempt at a solution

[tex]\rho=\dfrac{Mass}{volume}[/tex]

so

[tex]1.3=\dfrac{500}{1600}(\dfrac{273}{273+\Delta T})[/tex]

[tex]4.16=\dfrac{273}{273+\Delta T}[/tex]

[tex]1135.68 + 4.16\Delta T = 273[/tex]

[tex] \Delta T = -207.375[/tex]

But then this means the temperature change is negative which can't be right... Can anyone help please?

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# Homework Help: Density and Pressure Questions

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