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We have an old freezer. I believe the thermostat is broken and think it may run all the time. Food in it gets really cold (Ice cream like concrete). I've tried measuring the temp with a freezer thermometer, but the needle is well below the lowest temperature (-20F or -30F). The digital thermometer I have just shows "Error".

Aside from buying an expensive specialty thermometer (if they even exist), can anyone suggest the best straightforward way to calculate the approximate temp without any special equipment?

I was thinking of something like putting a solid metal object (to avoid phase changes) in the freezer for a day or two and then placing it in a known quantity of liquid of a time and measuring the temp of the liquid (or the object) after a set time (measuring the mass of everything of course). There may be an entirely better approach.

I'm not shooting for super accuracy, so not looking to build a super insulated rig. Just looking for a reasonably simple method to get a decent estimate (within 5 degrees?). Even specific suggestions on my proposed method appreciated.

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    $\begingroup$ A kitchen freezer should run at 0⁰F. If yours is below that, it's broken. Your question about a home thermometry experiment may be on-topic for us, but the practical question of fixing your appliance is not. Consider editing to clarify. $\endgroup$
    – rob
    Nov 29, 2023 at 21:34
  • $\begingroup$ Is your goal to fix/diagnose your refrigerator or measure temperature? $\endgroup$
    – Jagerber48
    Dec 5, 2023 at 18:53
  • $\begingroup$ Measure, not fix. New one is on order! $\endgroup$
    – PStallings
    Dec 6, 2023 at 21:54
  • $\begingroup$ I doubt that the freezer is capable of producing temperatures below $-20° \rm F$, though I could be wrong there. The far more likely scenario is that your temperature sensor is simply broken. $\endgroup$
    – RC_23
    Dec 27, 2023 at 2:20

5 Answers 5

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The freezing point of water ethylene glycol mixtures depends in the concentration of ethylene glycol. You can find data for this on the Engineering Toolbox web site (a valuable source of all sorts of info!). I did a quick graph to show the freezing point depression:

FP depression

So all you have to do is fill a few jars with different concentrations of ethylene glycol, put then in the freezer and see which ones freeze. I'd suggest using a binary search i.e. start with 30% and 60%, then if only the 30% freezes try 45% and 60%, or if neither freeze try 15% and 30%. And keep going narrowing the range until you have the accuracy you want.

Ethylene glycol is just anti-freeze as available at your local garage. Make sure you get neat ethylene glycol though and not the prediluted form.

A brief note on the toxicity of ethylene glycol since a few people have mentioned it, ethylene glycol has roughly the same toxicity as ethanol so it's hardly a major poison. If it was we wouldn't use it in our cars. However unlike ethanol it has a sweet flavour so it's more tempting to drink, especially for children. If you're going to put it in your freezer make sure it's well labelled and keep children away.

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    $\begingroup$ Would it be reasonable to start with a high concentration and instead add water until it freezes? $\endgroup$
    – Ezekiel
    Nov 30, 2023 at 16:16
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    $\begingroup$ @Ezekiel It depends on the trade off between doing the experiment quickly and saving on ethylene glycol. You probably need to leave the samples in the freezer at least overnight, so doing one sample at a time will be slow. On the other hand making up a dozen different samples and putting them in at the same time is more work and uses more glycol. But glycol is cheap! $\endgroup$ Nov 30, 2023 at 16:34
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    $\begingroup$ Is ethylene glycol more available/less expensive than propylene glycol? I'm not sure if it's a big deal, but propylene glycol is much less toxic and can do similar enough temp ranges that it should also work. $\endgroup$
    – JMac
    Nov 30, 2023 at 17:45
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    $\begingroup$ The physics of the answer is legit, but I would advise to put in the freezer a sweet toxic colorless substance. That freezer seems to be used for food (at least ice cream) and ethylene glycol is toxic and can be confused with something sweet and edible. $\endgroup$
    – Pere
    Nov 30, 2023 at 20:27
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    $\begingroup$ @Pere I assume there's a missing "not"? Also, once you're done with it, simply pouring it down the drain is probably illegal. $\endgroup$ Dec 1, 2023 at 0:41
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Mass of polystyrene cup + lid (with hole for thermometer) = $m_{\rm c}\,\rm g$ (assume thermal capacity is negligible) Mass of polystyrene cup + lid + water = $m_{\rm c+w}\,\rm g$
Mass of water = $m_{\rm w} = m_{\rm c+w} -m_{\rm c}\,\rm g$
Initial temperature of water = $\theta_{\rm i}\,^\circ \rm C$
Specific heat capacity of water, $c_{\rm w} = 4.2\,\rm J\,g^{-1}\,^\circ C^{-1}$

Mass of metal object of known composition = $m_{\rm m}\,\rm g$
Metal specific heat capacity, $c_{\rm m}\,\rm J\,g^{-1}\,^\circ \rm C^{-1}$
Keep metal in freezer for a day

Add to metal to polystyrene cup, close lid, shake without spillage until temperature constant, $\theta_{\rm f}\,^\circ \rm C$

Freezer temperature = $\theta\,^\circ \rm C$
$m_{\rm m}\,c_{\rm m}(\theta_{\rm f}-\theta) = m_{\rm w}\,c_{\rm w}(\theta_{\rm f}-\theta_{\rm i})$

An experiment that you might have done in your youth?


Data from @PStallings
This is the method I tried. I used a 1.5" chrome steel ball bearing as the thermal mass and 150 g of water at 68F in a styrofoam cup.
First time water froze almost immediately. Retried with 80 proof vodka and more mixing. At 1 min vodka at 56F.
At 2 min, at 55F. Same for next 3 min. Then slowly climbed for next half hour.
Converting to metric, vodka went from 20c to 13c. Change of 7c.
Specific heat ~2.4 gives 150g
2.4*(7c) = 2520. Mass of steel = 226g. Specific heat ~ 0.466.
2520/(2260.466) = 24. Leaves us with 13c-Initial Steel=24.
Gives -21c = -5.8F. Warmer than expected!

I checked the data assuming that the ball bearing was made of steel with a density of $7.85\,\rm g/cm^3$ which gives a mass for the ball bearing of $227\,\rm g$ which is in agreement with the OP.
However given that $80$ proof is equivalent to $40\%$ ethanol this table lists the specific heat capacity of the alcohol as $4\,\rm J/g\,K$.
Making the conversions to degree Celsius with $\theta$ as the initial temperature of the ball bearing the energy balance equation is, $150 \times 4 \times (20-12.8) = 227\times 0.466\times (12.8 - \theta)\Rightarrow \theta = -28^\circ \rm C$.

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    $\begingroup$ So, you propose measuring water temperature change induced by the metal, using the ratios of their masses and heat capacities? Be careful about freezing: the heat capacity of freezing is huge, and the water will freeze onto the metal. And you'll bleed some heat while getting the heat to transfer I think. $\endgroup$
    – Yakk
    Nov 30, 2023 at 17:24
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    $\begingroup$ Basically you're proposing to set up a bomb calorimeter $\endgroup$
    – Dave
    Nov 30, 2023 at 19:08
  • $\begingroup$ @Yakk well, I think the vigorous mixing step is supposed to sort all that out. $\endgroup$ Nov 30, 2023 at 22:50
  • $\begingroup$ @Yakk The metal object is assumed not to have such a large thermal capacity that the water in the polystyrene cup freezes. As well as shaking the cup a way to try and make sure that does not happen is to have warm/hot water in the cup. $\endgroup$
    – Farcher
    Dec 1, 2023 at 0:08
  • $\begingroup$ This is the method I tried. I used a 1.5" chrome steel ball bearing as the thermal mass and 150 g of water at 68F in a styrofoam cup. First time water froze almost immediately. Retried with 80 proof vodka and more mixing. At 1 min vodka at 56F. At 2 min, at 55F. Same for next 3 min. Then slowly climbed for next half hour. Converting to metric, vodka went from 20c to 13c. Change of 7c. Specific heat ~2.4 gives 150g*2.4*(7c) = 2520. Mass of steel = 226g. Specific heat ~ 0.466. 2520/(226*0.466) = 24. Leaves us with 13c-Initial Steel=24. Gives -21c = -5.8F. Warmer than expected! $\endgroup$
    – PStallings
    Dec 20, 2023 at 2:00
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Buy a small bottle of 100 mL vodka (white distilled spirits). This is an alcoholic beverage with 40% concentration of ethanol by volume, and it freezes at around -27 degrees Celsius. If it does not freeze, you can later dilute the vodka with a known volume of distilled water to obtain varying concentrations, and compare with lookup tables on the Internet. You will need a tool like the small plastic syringe from the pharmacy to perform the required volumetric measures.

Alternatively, if the vodka somehow freezes, you can buy the stronger 95% rectified spirits and dilute it to obtain concentrations above 40%, but most home freezers aren't that cold.

You can also get an immediate, rough estimate by just making a saturated table salt solution and putting it in the freezer. The freezing point is around -21 degrees Celsius, so you will know if you are above that or below.

Do not put a poison like ethylene glycol in a place where you store food, ever! This is beyond stupid and just asking for trouble.

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Adding a known amount of energy via a microwave oven

Pour a known volume/mass of water into a lightweight plastic cup. Let's assume you pour 100 ml = 100 g.

Place the cup into the freezer and allow it to thermalize and freeze. Then place the cup into a microwave oven with and run it until the ice completely melts. Note the run time and measure the final temperature.

Let's assume it took 2 minutes = 120 seconds and the end temperature is $65.0\ ^\circ \mathrm{C}$.

Find the IEC rated power of the microwave, which is the commonly advertised value. Let's assume 1200 W. This is critical: the IEC wattage rating is based upon the temperature change of a test cup of water, which is perfect for us. However, it has a 2x factor due to historical reasons1. Therefore, your "1200 W" microwave delivers 600 W to the water.

In our example, we delivered $E=120\ \mathrm{s} \times 1200/2\ \mathrm{J/s} = 72 \ \mathrm{kJ}$ to the water.

Since the ice melted, we subtract the heat of fusion, $334\ \mathrm{J/g}$ or $33.4\ \mathrm{kJ}$ for our 100 g of water.

That leaves us with $38.6\ \mathrm{kJ}$ of heating for our 100 g mass, or $386\ \mathrm{J/g}$.

The specific heat capacity of water we assume is constant at $4.19\ \mathrm{J/g\ ^\circ C}$, so our temperature change $\Delta T = \frac{386\ \mathrm{J/g}}{4.19\ \mathrm{J/g\ ^\circ C}}= 92.2\ ^\circ \mathrm{C}$

Since the final temperature was $65.0\ ^\circ \mathrm{C}$, the initial temperature was $65.0-92.2=-27.2\ ^\circ\mathrm{C}$

The major simplifications here are that the density and heat capacity of water are unaffected by temperature, the IEC wattage rating is accurate and applies to the cup of water, and the cup's thermal mass and heat transfer with the environment is negligible.

We could make the experiment more accurate by measuring the effective microwave oven wattage ourselves, by finding the $\Delta T$ of an identical room-temperature test mass of water.


1 Historically, magnetron tubes were rated by the DC input power, as this was easier to measure than high power RF, and they were roughly 50% efficient. The IEC measurement method keeps this factor for continuity in advertising.

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    $\begingroup$ This would work in terms of energy added to the system, but one additional assumption being made here is that ice absorbs energy well in the microwave range, and sadly for this experiment this is not the case. Ice is not a good microwave candidate: physics.stackexchange.com/a/136078/44198 $\endgroup$
    – Yos233
    Nov 30, 2023 at 20:22
  • $\begingroup$ @Yos233 That doesn't actually apply, there's faulty reasoning there. Try melting ice in your microwave oven. It works fine. Why? A microwave oven isn't free space where the energy makes one pass, it's a cavity where energy is confined or "bounces around". The energy has to go somewhere, which is either absorbed in the ice/water, lost resistively the walls or back in the magnetron tube. Walls are adequately electrically conductive so it's not a big factor (touch the walls). Magnetron losses, kind of, but the magnetron itself is obviously a low-loss structure. $\endgroup$
    – user71659
    Nov 30, 2023 at 20:34
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Take some liquid, put 10% of it in the freezer, wait a few hours. Measure the temperature of the liquid you left out, then pour in the liquid from the freezer, and measure again. The temperature of the freezer is ten times the second temperature, minus nine times the first temperature (using an absolute zero scale, of course). You can adjust the proportions; if you put percentage $p$ in and leave $1-p$ out, it will be $(T_f-(1-p)T_i)/p$.

This depends on the liquid not having a phase change. John Rennie suggests using ethylene glycol, but that is somewhat expensive and toxic; most jurisdictions have regulations regarding disposing of it. Sugar doesn't depress the freezing point quite as much, but it's much cheaper and less toxic. You can also try using salt. Many artificial sweeteners are more effective than sugar, but also are more expensive.

Another option is to see if you can find high precision resistance tester. A decrease of 5 degrees Fahrenheit will decrease the resistance of a copper wire by a bit over 1%.

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