Genuinely curious now, let's see how much i remember of gen chem and how many assumptions I can get away with making...
First, let's assume that the tea acts and behaves like water.
Typically, the contents of an aerosol are stored at 2–8 times normal atmospheric pressure
According to FEA (European Aerosol Federation) standards, maximum pressures for typical aerosol cans are in the range 12–18 atmospheres.
Let's assume the can starts at 8atm and 24ºC (room temperature) and bursts at 18atm. What temperature will the can contents be when it bursts?
The ideal gas law, which states that pV=nRT (pressure * volume = amount of gas * ideal gas constant * temperature), allows us to model changes in pressure, temperature, and volume with the ratio P1 * V1 / T1 = P2 * V2 / T2.
Volume may not remain constant, since water expands by a factor of 1600 when it turns into steam. This process happens at 100ºC at a pressure of 1atm, 170ºC at 8atm, and 205ºC at 18atm, so we can confidently say that if we reach the point where any of the tea is converted into tea vapor, the tea will be sufficiently hot. So for simplicity's sake, let's assume a constant volume of 300mL.
Since the temperature of the tea contents far exceeds the boiling point of tea under pressure, we can assume that when the can explodes the tea will be at its boiling point, i.e., hot as shit.
I assume because you can add boiling water to the concentrate and skip the cooling that happens as it steeps so you can burn your mouth if that's your fetish
how could it possibly be 'hotter' you cant heat a fucking pressurised can
well, you could, but depending on how much pressure it's under it might not get all that hot before you get impaled by the can shrapnel
race between my tepid tea concentrate and the explosive limit of my aluminium can
Genuinely curious now, let's see how much i remember of gen chem and how many assumptions I can get away with making...
First, let's assume that the tea acts and behaves like water.
Let's assume the can starts at 8atm and 24ºC (room temperature) and bursts at 18atm. What temperature will the can contents be when it bursts?
The ideal gas law, which states that pV=nRT (pressure * volume = amount of gas * ideal gas constant * temperature), allows us to model changes in pressure, temperature, and volume with the ratio
P1 * V1 / T1 = P2 * V2 / T2.Volume may not remain constant, since water expands by a factor of 1600 when it turns into steam. This process happens at 100ºC at a pressure of 1atm, 170ºC at 8atm, and 205ºC at 18atm, so we can confidently say that if we reach the point where any of the tea is converted into tea vapor, the tea will be sufficiently hot. So for simplicity's sake, let's assume a constant volume of 300mL.
Since the temperature of the tea contents far exceeds the boiling point of tea under pressure, we can assume that when the can explodes the tea will be at its boiling point, i.e., hot as shit.
So what your saying is the hotter claim may be true but the tastier claim is dubious because you won't be tasting much of anything at this point.
It is very hard to taste how bad this tea is when your tongue is melted, yes
P1 = 8atm V1 = 300mL T1 = 24°CAt this state the tea is still a liquid, right? So the ideal gas law would be inappropriate.
mmmm fair. The ideal gas law would still apply to the NO2 or CO2 used to pressurize the can though
i encourage any anglo mad enough to buy this product to put this math to the test
I assume because you can add boiling water to the concentrate and skip the cooling that happens as it steeps so you can burn your mouth if that's your fetish