/sci/ - Science & Math

bout 3 fiddy

https://www.engineeringtoolbox.com/energy-storage-water-d_1463.html

Q=cm∆t

2 Liters

>>12003042
this is too hard

30cm

>>12003801
no, you just have a low attention span. It took me a while, read it carefully. If I, a pajeet whose first language isn't English can do it, so can you

>>12003801
>>12004357
perhaps I should expand. Look at the green line, and the overall volume 2L. The graph shows you how much energy is stored in the unit of your choice, here, I'm looking at SI, so kJ. You have the specific heat of water, identify the equilibrium point now that all three (dt, CP and Vtot) quantities are known. That will give you the temperature.

>>12004366
>>12003801
also, this only works for isothermal systems.

>>12003801
you have to be 18 to use this site

>>12002782
30°C

>likes berzum
>scientifically illiterate
makes sense

>>12002782
>What is the resulting temperature
Not as hot as your mother.

>>12003801
Just look at the volume-energy plots, they're all linear, meaning the "density" of the energy doesn't change if you change the amount of water you have, so 2L of water at the same temperature as some other 1L hold two times the energy.
Mixing 1L of 10°C water and 1L of 50°C water results in 2L of 30°C water. (because you have the same quantity of water, all you have to do is find the middle ground between their two temperatures, e.g. for 10°C and 50°C you get 10 + (50-10)/2 which equals 30)

>>12004573
Actually, the middle ground is much easier to calculate, shouldn't have complicated it like that, just find the average: (10+50)/2 = 30

>>12003250
ENTROPY BITCH

>>12004573
Thanks.
What about 1L of 0°C water mixed with 2L of 30°C water?

>>12005655
Maybe you can divide 60 liter-degrees by 3 liters.

>>12005678
Okay so it's that easy? Why did you mention some abstract Joules then if they're superfluous, smartass? To make your ego grow?

>>12005655
(1L * 0°C + 2L * 30°C) / (1L + 2L) =
60/3 °C =
20°C

>>12005685
I'm not the same person, and I was just assuming that energy is linear with temperature as well as volume.
Imagine that a liter of water of x degrees has z+kx joules of energy.
(So z is the amount of energy in a liter of water at zero degrees, and every additional degree adds k joules.)

Then if you have amount a at x degrees, and amount b at y degrees, the combined energy in joules will be a(z+kx) + b(z+ky).
Which is (a+b)z + akx+bky =
(a+b)z + k(ax+by)

The joules per liter of the mix will be [(a+b)z + k(ax+by)]/(a+b) =
z+k(ax+by)/(a+b), which can be converted back to degrees per liter by subtracting z and dividing by k:
(ax+by) / (a+b)

>>12005795
sorry, at the end I should have said converted back to degrees, not degrees per liter.