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.

[–] fastandcurious@lemmy.world 19 points 10 months ago* (last edited 10 months ago) (3 children)

Who are ye, so wise in the ways of Thermodynamics

[–] can@sh.itjust.works 5 points 10 months ago

Ye are rapper

[–] MeetInPotatoes@lemmy.ml 2 points 10 months ago

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.

[–] SoggyBread@lemmy.world 3 points 10 months ago (2 children)

Its going from 7 to 77, not 57

[–] airbreather@lemmy.world 23 points 10 months ago

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.

[–] RubberElectrons@lemmy.world 2 points 10 months ago

Oh I missed that 2 7, thought it was Delta 70 as well on first glance.

[–] platypus_plumba@lemmy.world 1 points 10 months ago* (last edited 10 months ago) (1 children)

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.

[–] prime_number_314159@lemmy.world 3 points 10 months ago (1 children)

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.

[–] platypus_plumba@lemmy.world 1 points 10 months ago* (last edited 10 months ago)

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!