Extra heat gets added out of nowhere when using Thermo Aquatuner

Problem Statement

I decided to test my understanding of heat exchange and temperature handling in Oxygen Not Included. So I used Sandbox and Dev Mode to set up a simple heating loop and then tried to calculate the resulting temperature of the liquid I was heating.

Alas, the lab test results diverged from my calculations by a lot. I would appreciate any help finding flaws in my reasoning.

Heating

All the materials inside the box started at 10.0 ℃.

Then I fed 4 kg worth of Super Coolant at -35.1 ℃ to the Thermo Aquatuner.

As expected the coolant got cooled by exactly -14.0 ℃ down to -49.1 ℃.

Calculations and Expected Result

Insulated Tiles of the box are made of Insulation which has Thermal Conductivity of 0.0. This means the box itself cannot heat up or cool down, it also perfectly insulates its contents preventing any heat exchange between the insides and the outsides.

As an extra measure to prevent unwanted heat exchange, the area outside of the box is vacuumed.

So, since neither the box itself nor the vacuum outside of it could exchange heat I assumed that all the heat "extracted" from the coolant should go to the insides of the box.

Here's the inventory of the box contents:

Insulated Liquid Pipes are made of Insulation, so they don't participate in the heat exchange.

For the reference, here are the properties of the materials exchanging heat:

Material Specific Heat Capacity, DTU/g/℃ Mass, kg
Water 4.179 2
Polluted Water 4.179 2
Thermium 0.622 1225
Super Coolant 8.440 4
1. Heat required to heat up an object is the product of material's heat capacity and temperature change:

ΔQ = C × ΔT

2. Heat capacity is the product of specific heat capacity and mass of the material:

C = c × m

3. Temperature change is the product of specific heat capacity and mass of the material:

ΔT = Tfinal − Tinitial

4. Given the formulae above the amount of heat given by the coolant is this:

ΔQsuper coolant = c × m × (Tfinal − Tinitial)

ΔQsuper coolant = 8.440 × 4 × (−49.1 − (−35.1))

ΔQsuper coolant = 8.440 × 4 × (−14)

ΔQsuper coolant = −472.64 kDTU

Note, the result is negative because the coolant is cooled instead of heated.

Also note, it's kDTU not DTU since mass is in kg not in g.

5. Since the system is closed (no heat is received from the outside, none is dissipated either), the heat amount stays the same, in other words the heat change is zero:

ΔQsuper coolant + ΔQbox contents = 0

ΔQbox contents = −ΔQsuper coolant

6. Assuming we give the system enough time to stabilize, the initial temperature of the box contents is going to be the same for all its parts such as pipes, wires, liquids, devices, etc. The same applies to the final temperature.

Therefore we can treat the box contents as a single pseudo-object that has some heat capacity, so having formula `(1)` in mind we get the following:

ΔQbox contents = Cbox contents × ΔTbox contents

7. Heat capacity of the box contents is the sum of heat capacities of its parts:

Cbox contents = Cthermo aquatuner + Cconductive wire + Cwater + Cpolluted water

Cbox contents = mthermo aquatuner × cthermium + mconductive wire × cthermium + mwater × cwater + mpolluted water × cpolluted water

Cbox contents = (mthermo aquatuner + mconductive wire) × cthermium + mwater × cwater + mpolluted water × cpolluted water

Cbox contents = (1200 + 25) × 0.622 + 2 × 4.179 + 2 × 4.179

Cbox contents = 1225 × 0.622 + 2 × 4.179 + 2 × 4.179

Cbox contents = 761.95 + 8.358 + 8.358

Cbox contents = 778.666 kDTU/℃

8. Now starting from the formula `(6)` and substituting formulae `(3)`, `(5)`, `(4)` and `(7)` we can derive the final temperature of the box contents:

ΔQbox contents = Cbox contents × ΔTbox contents

ΔTbox contents = ΔQbox contents ÷ Cbox contents

Tbox contents final − Tbox contents initial = ΔQbox contents ÷ Cbox contents

Tbox contents final = ΔQbox contents ÷ Cbox contents + Tbox contents initial

Tbox contents final = −ΔQsuper coolant ÷ Cbox contents + Tbox contents initial

Tbox contents final = −ΔQsuper coolant ÷ Cbox contents + Tbox contents initial

Tbox contents final = −(−472.64) ÷ 778.666 + 10.0

Tbox contents final = 472.64 ÷ 778.666 + 10.0

Tbox contents final ≈ 10.6℃

So the conclusion is that after the box contents get heated and all its contents parts take their time to balance the temperature the final temperature of the box contents would be 10.6℃.

Actual Result

And here's the actual results that I got in the game:

The actual temperature is 12.8℃ instead of 10.6℃ calculated above.

I would appreciate any clues as to where I made a mistake in my calculations.

• Hi Ihor, great first question! Welcome to Arqade! Commented Sep 23, 2023 at 14:14
• Out of curiosity, are the results the same if you send in a single 4 kg packet of Super Coolant, or 2*2kg packets? Just trying to find ways to find what the variable might be. Commented Sep 26, 2023 at 20:53
• @SimonForsberg, just tried packets of different sizes: starting from 10 kg and down to 100g each. Turns out, the size of the "pieces" doesn't matter as long as the "whole" stays the same. At least that makes sense to me. Commented Sep 27, 2023 at 18:53