Heat capacity of lava per degree Celsius per unit of mass
Multiplied by temperature differential vs ambient, multiplied by mass = total extra heat energy
Then you calculate the sum of heat capacity multiplied by mass for lava and for water, and calculate from that how many degrees above ambient the two masses will land at when combined as the extra energy above is divided over both (assuming water starts at ambient temp)
It won’t be exact because heat capacity varies in materials as temperature changes, both steam and solidification of lava (state change) will contribute significantly, but it’s a decent first estimate
Okay. Then they add more and it will boil quickly. I guess the question boiles (huehue) down to how much water you can turn into stream per amount of lava or the inverse, how much lava you can cool down per amount of water.
The phase change from liquid water to stream will, by the way, not just contribute significantly but be by far the majority of energy needed. Simply heating water up, ignoring the phase change and changes of the heat capacity, with the same energy as it takes to go from liquid to gas (2257 kJ/kg) would result in a temperature rise of…
dT = 2257 kJ/kg / 4.2 kJ/(kg*K) = 537 K
Assuming enough water that most of it doesn’t boil, then my math would still check out, but yeah, any substantial amount of boiling forces you do do the math in multiple steps to handle that
Bruh. Someone else on this thread has already clarified to you the easy and what I was expecting question: what happens to the water and lava in the water bucket.
You already answered that question in this thread.
Do what math? I honestly don’t know what you guy’s actually expect it to look like, so I don’t know where to start explaining.
Heat capacity of lava per degree Celsius per unit of mass
Multiplied by temperature differential vs ambient, multiplied by mass = total extra heat energy
Then you calculate the sum of heat capacity multiplied by mass for lava and for water, and calculate from that how many degrees above ambient the two masses will land at when combined as the extra energy above is divided over both (assuming water starts at ambient temp)
It won’t be exact because heat capacity varies in materials as temperature changes, both steam and solidification of lava (state change) will contribute significantly, but it’s a decent first estimate
Okay. Then they add more and it will boil quickly. I guess the question boiles (huehue) down to how much water you can turn into stream per amount of lava or the inverse, how much lava you can cool down per amount of water.
The phase change from liquid water to stream will, by the way, not just contribute significantly but be by far the majority of energy needed. Simply heating water up, ignoring the phase change and changes of the heat capacity, with the same energy as it takes to go from liquid to gas (2257 kJ/kg) would result in a temperature rise of… dT = 2257 kJ/kg / 4.2 kJ/(kg*K) = 537 K
Assuming enough water that most of it doesn’t boil, then my math would still check out, but yeah, any substantial amount of boiling forces you do do the math in multiple steps to handle that
Seriously? The lava in water math.
It’s high school stuff if you bother to look up the specific heat and make some reasonable guesses.
It’s masters degree in thermal fluids engineering math
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No it’s not. Google specific heat of lava. Shit’s been done already.
And not just for academics. Even Randall Monroe. He’s smart, but he’s not an academic.
Mate, I model thermal processes for a living. The question is: What is the question? What to calculate? What expectations are there to (dis)proof?
Bruh. Someone else on this thread has already clarified to you the easy and what I was expecting question: what happens to the water and lava in the water bucket.
You already answered that question in this thread.
“Bruh”, everyone knows what happens, we are looking at it in the video. Lava gets cold, water hot. Obviously. There is not much to go on about.
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