Every 1 minute, 3L of water gets heated by 50K. With a specific heat capacity of 4200 J / kg / C and density of 1 g / mL, it takes 3 x 4200 x 50 = 630000J per minute.
With a rate of 4.0 x 10^4 J / g for the heater, we can get the rate of combustion with 630000 / 40000 = 15.75 g per minute.
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You have 3 liters of water heating up by 50 degrees celsius. It takes 4184 joules to heat 1 liter of liquid water by one degree Celsius, so it takes 627600 joules to heat 3 liters by 50 degrees. Dividing by 40000 joules per gram of fuel, it will take 15.69 grams of fuel per minute. Finally, for significant digits, we have to round to 16. grams of fuel per minute.
Edit: for most sciency uses, 1.6 times 10^1 grams of fuel per minute is likely the preferred way to write that.
Who are ye, so wise in the ways of Thermodynamics
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Probably a witch, and we would normally burn them…but burning a witch who is wise in the ways of Thermodynamics is like shooting marsh water at a swamp monster.
Its going from 7 to 77, not 57
Its going from 7 to 77, not 57
“from 2 7°C to 77°C” is either “from 27°C to 77°C” with an extremely problematic line break position, or something unintelligible.
Oh I missed that 2 7, thought it was Delta 70 as well on first glance.
Hey, just curious… when you say that it takes a specific amount of joules to heat water, why is the time variable, surface area of the body of water and surrounding temperature ignored? It seems so weird to me to hear that it takes a fixed amount of energy to heat water from temperature A to temperature B. I feel there are so many more variables involved.
Are you just ignoring variables for the sake of being able to give an answer? Seems to me like a classic scenario of “they taught me to ignore these variables in physics class”. Or am I wrong? I’m very curious. I was trying to solve it logically but it wasn’t possible due to missing variables. I have no clue about thermodynamics.
I’m ignoring many factors for the sake of being able to answer. There are some kinds of heating, especially using burning fuels that are nearly 100% efficient, but we don’t know why it needs to get to that temperature, or how long it needs to stay that hot - so even if the transfer of heat is 100% efficient, this computation may underestimate the actual needs.
Ah, alright, I was feeling dumb because the formulas made no sense to me but it is just a model providing a practical approximation. Thanks!
I believe that should be 10 to the 4th power at the end there.
takes off nerd glasses
Thanks for reminding me why I ended up at a third-rate college.
The answer is 7.
Sorry, my mind goes blank when see that fool.
