# Question about calculating the minimum temperature in hot air balloons

• ianc1339
In summary, the conversation discusses calculating the weight of the air displaced during take off using the Archimedes principle and determining the weight of the air inside the balloon for equilibrium. The relationship between density and temperature is also mentioned, with the final conclusion being that the temperature inside the balloon just before take off is 376.6K.
ianc1339
Homework Statement
A hot air balloon has a volume of 2500 m^3 and the fabric and the passengers have mass of 800 kg. It is taking-off when the air around the balloon has a temperature of 10 degrees Celsius and a density of 1.29 kgm^-3. Calculate the minimum temperature of the air inside the balloon for take off. Note that the pressure inside and outside the balloon must be equal, since the balloon is open at the bottom.
Relevant Equations
pV=nRT
pV=1/3Nm<c^2>
<E_k>=1/2m<c^2>=3/2kT
p = 1/3 x density x <c^2>
First, I tried using the Archimedes principle and calculated the weight of the surrounding air displaced when taking off.

##W = 2500\times 1.29\times 9.81 = 31637.25 N##

But then, I got stuck and do not know how to proceed from here on.
I don't want the full solution yet but can I get some guidance/hints as to how to approach this question?

Thanks

etotheipi said:
What is the buoyant force on the balloon, in terms of its variable volume ##V##?

Is it ##\rho V g##?

Apologies, I misinterpreted the question. The volume is most definitely not variable!

You have the balloon at that given volume, and you know the buoyant force and the weight of the material. What must be the weight of the air inside for equilibrium? How might we then relate density and temperature?

Last edited by a moderator:
Lnewqban and ianc1339
etotheipi said:
Apologies, I misinterpreted the question. The volume is most definitely not variable!

You have the balloon at that given volume, and you know the buoyant force and the weight of the material. What must be the weight of the air inside for equilibrium? How might we then relate density and temperature?

So the weight of the material is ##W=mg=800\times 9.81 = 7848N##. So the weight of the air inside must be ##31637.25 - 7848 = 23789.25N##.

Since volume is proportional to temperature, as temperature increases, density decreases. So temperature is inversely proportional to temperature?

ianc1339 said:
So temperature is inversely proportional to temperature?

Careful , though I think you meant to say density, which is indeed correct. You can show it from the good old ideal gas law ##PV = nRT \implies P = \frac{n}{V} RT##. Scale ##n## by the molar mass and you get a different constant of proportionality, but essentially ##\rho \propto \frac{1}{T}##.

You're already given one pair of values for density and temperature, so can you work out the new temperature? You could write out the proportionality and find the constant, or you could equate some 'conserved' quantity.

Last edited by a moderator:
Lnewqban and ianc1339
So the relation becomes ##\frac{1}{T}=\frac{R}{p}\times\rho## where ##\frac{R}{p}## is a constant.
##\frac{R}{p} = \frac{1}{T\rho} = 2.7377\times 10^{-3}##.

Since the weight of the air inside the balloon before take off is ##23789.25N##, the density of the air inside the balloon is ##\rho = \frac{m}{V}=\frac{W/9.81}{V} = \frac{23789.25/9.81}{2500} = 0.97kgm^{-3}##

So the temperature inside the balloon just before take off is ##\frac{1}{T}=2.7377\times 10^{-3}\times 0.97kgm^{-3}\\ \therefore T = 376.6K##

And yes, this was the answer (The answer stated 380K rounded to 2 s.f.).

Thank you so much for your help!

Lnewqban and etotheipi
ianc1339 said:
So the relation becomes ##\frac{1}{T}=\frac{R}{p}\times\rho## where ##\frac{R}{p}## is a constant.
##\frac{R}{p} = \frac{1}{T\rho} = 2.7377\times 10^{-3}##.

This is essentially correct, just note that the ##R## in that equation is the so-called "specific" gas constant, i.e. just divided by the molar mass: ##R_{\text{specific}} = \frac{R}{M_{r}}##.

ianc1339 and Lnewqban

## 1. How is the minimum temperature calculated in hot air balloons?

The minimum temperature in hot air balloons is calculated using the Ideal Gas Law, which states that the pressure, volume, and temperature of a gas are all directly proportional. This means that as the temperature decreases, the volume of the gas also decreases, causing the balloon to sink. Therefore, the minimum temperature is calculated by finding the temperature at which the volume of the gas inside the balloon is equal to the volume of the surrounding air.

## 2. What factors affect the minimum temperature in hot air balloons?

The minimum temperature in hot air balloons can be affected by several factors, including the volume of the balloon, the weight of the payload, the altitude, and the outside air temperature. These factors can all impact the amount of hot air needed to maintain lift and can affect the minimum temperature required to keep the balloon afloat.

## 3. Can the minimum temperature change during a hot air balloon flight?

Yes, the minimum temperature can change during a hot air balloon flight. As the balloon ascends or descends, the outside air temperature and pressure can change, affecting the volume and density of the hot air inside the balloon. This can cause the minimum temperature to fluctuate and may require adjustments to the burner to maintain lift.

## 4. How does the minimum temperature affect the flight of a hot air balloon?

The minimum temperature is crucial for the flight of a hot air balloon as it determines the amount of hot air needed to maintain lift. If the temperature is too low, the balloon will sink, and if it is too high, the balloon will rise uncontrollably. Pilots must carefully monitor the minimum temperature to ensure a safe and stable flight.

## 5. Are there any safety precautions to consider when calculating the minimum temperature in hot air balloons?

Yes, there are several safety precautions to consider when calculating the minimum temperature in hot air balloons. Pilots must be aware of the weight limit of the balloon and ensure that the burner is not producing too much heat, which can cause the balloon to overheat and potentially catch fire. It is also essential to monitor weather conditions and avoid flying in strong winds or thunderstorms, as these can affect the minimum temperature and pose a safety risk.

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