null's math question

[wokelizard]

You need to know the masses of each given by the molar mass which is 55.85, 63.55 and 18.02 g/mol. Than means you have 55.85g of iron, 63.55g of cooper and 1802g of water.
You need to know the specific heat capacity of iron, copper and water which is 0.4605, 0.3768 and 4.182 J/gK. The heat gained by the water will equal the heat lost by the metals such that the temperature of all three is the same. Call it T. So 0.4605 * 55.85 * (200 - T) + 0.3768 * 63.55 * ( 200 - T ) = 1802 * 4.182 * ( T - 20 ). That's only got one unknown (T) so it's super easy to solve by collecting T and rearranging to give T = 21.18. It would be impossible to say what the temperature would be in 1 second without knowing the geometry of the system and how the heat moves. You would then have to apply the heat equation (a PDE) and the boundary conditions to solve.

 

[NoodleFucker3000]

Why are you asking the kiwis to do your chemistry homework Null

 

[Papa Adolfo's Take'n'Bake]

Dudes, this is nowhere near a grad school level problem requiring PDE's. This is autistically doing thermodynamic calculations on a shitposting forum.

Assume the iron and the copper are spheres lying in a sealed off Water container. Given the masses, we have about 1.8L of water total. These spheres thus have the following characteristics:

V(Fe)=7.096*10^-6 m^3
Aq(Fe)=1.7858*10^-4 m^2
h(Fe)=7.9 (W*(m^-2)*(K^-1))
k(Fe)=69.4 (W*(m^-1)*(K^-1))

Yielding
Biot(Fe)=0.0046

Similarly

V(Cu)=7.*108*10^-6 m^3
Aq(Cu)=1.7878*10^-4 m^2
h(Cu)=13.1 (W*(m^-2)*(K^-1))
k(Cu)= 392 (W*(m^-1)*(K^-1))

Yielding

Biot(Cu)=0.0013

Can we please not try to flex "muh super
spooky grad level math" over a system that is extremely well described by Newton's law of cooling? I know exactly 0 people that would do anything but just use newton's law to describe the system as follows.

The iron is now at 463.322 K (roughly 190 degrees centigrade.)

The copper is now at 456.919 K (roughly 184 degrees centigrade.)

Given the assumed system we are looking at, that heat went into the water and nowhere else, yielding the water going up by roughly 0.087 degrees centigrade, using the basic heat capacity calculation from earlier.

If you are curious how this shit usually looks and works, here: http://web.mit.edu/16.unified/www/SPRING/propulsion/notes/node129.html

 

[Polyboros2]

Man, no longer needing to know this kinda shit it's half the reason I graduated from college.

 

[WarJams]

All I can tell you is that steel is heavier than feathers.

 

[Banditotron]

I just use duct tape and scrap wood to build things so this is above my pay grade

 

[Rafal Gan Ganowicz]

https://www.youtube.com/watch?v=Liqs0k4bXvY

 

[Lumin]

Apologies if I'm missing something, but what "gameplay" are you working on, Null? Or is that too early to disclose? Hobby project or something more serious, I saw your post about how you're coding again so have fun with it.

 

[D_Tractor]

Heat transfer is a big part of the game Oxygen Not Included and apparently this is how they do it:

https://forums.kleientertainment.com/forums/topic/100121-heat-transfer-equations/

 

[MrTroll]

Nice try but I'm not going to help you make a bomb. Furthermore you have been reported to the FBI for posting this. Good day.

 

[Racist Trash]

Please stop enabling Null's hot Iron and Copper blend addiction we stole his recipe book for this very reason.

 

[Un Platano]

Dudes, this is nowhere near a grad school level problem requiring PDE's. This is autistically doing thermodynamic calculations on a shitposting forum.

Assume the iron and the copper are spheres lying in a sealed off Water container. Given the masses, we have about 1.8L of water total. These spheres thus have the following characteristics:

V(Fe)=7.096*10^-6 m^3
Aq(Fe)=1.7858*10^-4 m^2
h(Fe)=7.9 (W*(m^-2)*(K^-1))
k(Fe)=69.4 (W*(m^-1)*(K^-1))

Yielding
Biot(Fe)=0.0046

Similarly

V(Cu)=7.*108*10^-6 m^3
Aq(Cu)=1.7878*10^-4 m^2
h(Cu)=13.1 (W*(m^-2)*(K^-1))
k(Cu)= 392 (W*(m^-1)*(K^-1))

Yielding

Biot(Cu)=0.0013

Can we please not try to flex "muh super
spooky grad level math" over a system that is extremely well described by Newton's law of cooling? I know exactly 0 people that would do anything but just use newton's law to describe the system as follows.

The iron is now at 463.322 K (roughly 190 degrees centigrade.)

The copper is now at 456.919 K (roughly 184 degrees centigrade.)

Given the assumed system we are looking at, that heat went into the water and nowhere else, yielding the water going up by roughly 0.087 degrees centigrade, using the basic heat capacity calculation from earlier.

If you are curious how this shit usually looks and works, here: http://web.mit.edu/16.unified/www/SPRING/propulsion/notes/node129.html

I was going for a more general solution to the problem than approximating this one particular case since it seemed like Null was interested in mire than just the answer to this one particular problem. Your solution simplifies the equation because it has a low Biot number, and because the temperature of the water changes little. Those factors make it possible to approximate this as a Newton's law of cooling problem, as I have already addressed. But those approximations cannot be made for all systems, such as if a fluid with a lower heat capacity were used, or if the metal became very large in proportion to the fluid.

Given that this is for a game though I think that kind of accuracy is unnecessary, and that any sort of exponential decay would feel realistic enough to the end user.

 

[GogglyGoblin]

I think it'd be a terrible idea to want to simulate this kind of thing with great precision, unless the point of doing it is a programming exercise. I think you're better off equilibrate their temperatures using a simple linear function and have its slope change proportionally to the difference in temperature. I say this because the user will not notice any difference and you will save computing time.

 

[Yotsubaaa]

In my simulation I have 'jars' with different matter. They don't necessarily have shape. I was just hoping to do a basic simulation of some physics for the same of complementing gameplay later on.

I just think it'd be really shitty for this to be like, "okay there's now a hot ball in this 3x3 room, so now the ball and the air are all the same temperature immeduately."

What would be a cheap and good enough solution for heat dispersion?

The other Kiwis here have given you proper, mathematical answers, but I'll give you a pragmatic one: honestly, you might want to just fudge it a little if it's for your video game. Realistic heat effects don't always look as cool or as 'intuitive' in video games/animation as just using simple rate equations and linear interpolation.

 

[Picklechu]

I can tell you how to make the water vote Republican but that's it.

 

[XYZpdq]

put them all in your butt and they end up a toasty 98.6

 

[Distant Stare]

Copper - 63.546 grams per mole
Iron - 55.85 grams per mole
water - 18.015 grams per mole

100 mol times 55.85 g*mol^-1 = 5.585x10^3 grams

The specific heats
Copper - 0.384 joules*gram^-1 *kelvin^-1
Iron - 0.449 joules*gram^-1 *kelvin^-1
water - 1.865 joules*gram^-1*kelvin^-1 *

Get the total amount of energy in the system by adding up all the components
= Specifc_Heat * grams * temp (in K)
Copper - 1.154687x10^4 J
Iron - 1.18663x10^4 J
Water - 9.8361^5 J

Sum = 1.007x10^6 J in of energy in the system

Just assume that there is just water to make calculations easier. There is 1.8 Kg of water, compared to only a few grams of metal. We use theta = Q/mc) for changes in temp.

theta = ( 9.0609x10^4 J)/(1.8x10^3 g * 1.865 J*g^-1*K^-1)
theta = 2.999^2 kelvin
temp = 26.7 centigrade at most @Null. It should be a little less because we did not include the metal.

As for after one second, in real life there will be no immediate change. Once the metal hits the water it will flash boil and create an insulating layer of steam. Not to mention that the metal has very low surface area to transfer heat, and water is a good insulator.

 

[Return of the Freaker]

All I can tell you is that steel is heavier than feathers.

There both ah killgram

 

[Tiny Clanger]

42
Or Jesus

These are always the answer to everything.

 

[Un Platano]

equilibrate their temperatures using a simple linear function and have its slope change proportionally to the difference in temperature.

This is exactly what Newton's law of cooling does. It assumes that the rate of change of temperature is proportional to the difference in temperatures between the object and its environment at any given time.

dT/dt is the rate of change (slope) of the temperature, t is the time, T is the temperature of the object, Ta is the ambient temperature, and k is a thermal conductivity constant.

When you actually solve this for temperature T at any time t though, it's not linear anymore. This is a type of differential equation, but it's a separable differential equation, which are very easy to solve. We can put all the T's on one side, and all the t's on the other, and integrate both sides to eliminate the differential components, and we are left with a relatively simple expression of T in terms of t and some constants.

The e^c component that arises from this is the initial condition of the system and is solved using the known temperature at time t = 0, usually given as To - Ta. The absolute values bars that disappear at some point come from whether or not the body is increasing or decreasing in temperature, because the law works for both heating and cooling.

water - 1.865 joules*gram^-1*kelvin^-1 *

Your answer is off by about 5 degrees because you used the wrong specific heat for water. The specific heat of liquid water at standard conditions is about 4.2 J/gK. You used the specific heat of very low pressure steam by the looks of it.

You do have a good point about the insulating of effect of boiling at the surface. That's called the Leidenfrost effect, and it's a pain in the ass to calculate, so I'm not going to bother.