# Lake Ice Temperature

Given:

1. A lake with an established sheet of ice of some nominal thickness covering it.
2. It is night (no radiant energy from the sun).
3. No wind.
4. Air temperature -10 degrees Celsius.

Will the temperature of the exposed upper surface of the ice be approxumately the same as the air temperature while the submerged surface is perhaps no more than 0 degrees Celsius?

Or would the liquid water keep the upper ice surface slightly warmer than air temperature?

## 4 Answers

The answer is "it depends." Here are some of the factors on which it depends:

• The thickness of the ice.
Ice is a mediocre conductor of heat, about the same as rock. A thick layer of ice somewhat insulates the upper surface of the ice from the ~0 °C water just below the ice. A thin layer of ice, the ice will be at ~0 °C.
• The average wind speed.
Thermal conductivity is the dominant form of heat transfer in windy and breezy conditions. Radiative cooling dominates when the wind is very calm.
• The percent cloudiness, and the altitude of the clouds.
The percent cloudiness is a key driver in the night sky temperature (which is very important when radiative cooling dominates over thermal conductivity). Low clouds are warmer than higher clouds, so the height of the bottom of the clouds is important when clouds are present.
• The relative humidity.
Humidity is another key driver of the night sky temperature.
• The time of day (or time of night, in this case).
Air has a much lower heat capacity than does ice. Ice holds the warmth of the day much longer than does air.
• The amount of snow that is atop the ice.
Even a few centimeters of snow acts as a rather nice blanket. A meter of snow makes for a very, very nice blanket.
• What you mean by "air temperature -10 degrees Celsius."
The standard is to measure the temperature at 1.5 meters above the surface. I'll assume the standard.

The question eliminates one of the variables; there is no wind. It doesn't truly address the thickness of ice; the question stipulates ice of "some nominal thickness." There's a big difference between ice that is safe to walk on versus ice that is safe to drive a four wheel truck on versus ice that is safe to drive a fully-loaded eighteen wheeler. It doesn't touch on the other variables at all.

To illustrate, suppose the ice is covered with a meter of snow, and it's snowing. That meter of snow will make for a very nice blanket. That it's snowing means its cloudy and that the humidity is high. The surface of the ice will be just a bit below 0 °C.

On the other hand, suppose it's just before sunrise, the ice has no snow cover and is thick enough to support a four wheeled vehicle (~ 40 cm thick), the sky is perfectly clear, and the air is extremely calm and extremely dry. Now the surface of the ice can be well below -10 °C.

A related phenomenon occurs on clear, calm, and dry nights when the low is to 4 °C or so. You might well see frost on the roof of your house and on your grass, and if you have a pet with an outdoor water bowl, the water might well be covered by a thin layer of ice. Yet the temperature never dropped below freezing. The reason is radiative cooling dominates over thermal conduction on those clear, calm, and dry nights.

• Thanks for the feedback. Sorry I can't be more specific. The whole question arises from an encounter with a trainer of outdoor survival skills contending that it is warmer to camp and sleep on a frozen lake instead of nearby land because as he says, "ice is always 32 degrees". I am highly skeptical and wanted to learn more facts but there appear to be many variables involved that I had not considered. The claim that ice is always 32 degrees is what I am having trouble with. – Randy May 4 '15 at 18:32
• @Randy - An outdoor survival skills expert taught that? That's ludicrous. Most experts at winter survival teach that one should stay off of frozen lakes and rivers. Finding out that the ice is dangerously thin is life-threatening. – David Hammen May 5 '15 at 14:21
• Ice isn't always 32 degrees, what does he think Antartica is at 0C? – Suzu Hirose Sep 29 '16 at 12:48
• @SuzuHirose -- Who knows? The OP was not the person who related this point, so it's second hand information. To make matters worse, this was a drive-by question (the OP hasn't been back since the question was asked). On the other hand, this question is clearly about ice on a lake, not ice over Antarctica. – David Hammen Oct 3 '16 at 12:04

I measured ice temperature around March 4, 2017, on Yellowknife Bay, Great Slave Lake, Northwest Territories. The instrument was a new, but uncalibrated pocket infrared sensor by La Crosse Technology, Infra-Scan model, design range -27 to 390 deg F. This was for a story I was writing. Air temperature was -20 deg F, ice thickness was approximately 36 inches (visual estimate of clear ice). Conditions: overcast, approximately 2:00 p.m. Reading was on an area of bare ice, although there was snow over much of the bay, perhaps 10 inches deep. The clear area was about 20 feet in diameter. My readings ranged 8 deg to 14 deg F. The water beneath would be 39 deg F (max density) except at the interface where it would be approx 32 deg F, and conditions were such that the ice would continue to freeze for several more weeks. Night temperature ranged -35 deg to -40 deg F.

You would not have ice (at atmospheric pressure) if the temperature was about 0°C.

In your situation, the top of the ice is at -10°C and the bottom of the ice is at 0°C. Also, since there is very little water circulation, the bottom of the lake can be over 4°C.

• Since ice conducts heat two orders of magnitude more efficiently than air, it's not obvious to me that the top of the ice should be -10°C. Can you show your workings? – lemon May 3 '15 at 14:56
• @lemon What would prevent the surface of the ice from being the same temperature as the air? Any difference in temperature would quickly equilibrate. The interface would be at the temperature of the air. If the air becomes colder, then the heat from the water is transfered through the ice, making the ice thicker. – LDC3 May 3 '15 at 15:07
• Alternatively, a thin layer of air near the ice might take on the same temperature as the ice. That seems more likely since the flow of heat is in the upward direction. – lemon May 3 '15 at 15:17
• @lemon When the air get slightly warmer than -10°C, it will rise upwards, to be replaced by air at -10°C. So your thin layer of warmer air does not remain in place. – LDC3 May 3 '15 at 15:59
• If that's the case then why does the wind feel cold on my skin? – lemon May 3 '15 at 16:03

Since this problem is about temperature, we can use thermodynamics to solve it.

From Wikipedia, the Zeroth Law of Thermodynamics states:

If two thermodynamic systems are each in thermal equilibrium with a third, then they are in thermal equilibrium with each other.

Also, along with this is the fact that bodies in contact with each other will reach thermal equilibrium: heat flows from warmer to colder bodies.

With this in mind, if the air is in thermal equilibrium with the ice, and the water is in thermal equilibrium with the ice, then everything (air, ice, water) are in thermal equilibrium.

As an example, try putting a thermometer in a glass of ice water. It does not matter whether the thermometer is in direct contact with the ice, or in an area where it's mostly water, the entire glass of water is at 0$^{\circ}$C.

• Welcome on Physics SE and thanks for the contribution :) While what you wrote is thermodynamically totally sound, I think you need to consider that thermodynamics usually talks about equilibrium and has no concept of time. For two bodies to get into thermal equilibrium can take a lot of time ... – Sanya Nov 3 '16 at 9:48
• @Sanya Hello and thank you! Yes that's an interesting point. The Zeroth Law never specifically states a time variable does it? Can we assume that an imaginary small cube that surrounds air, ice, and water will be in equilibrium ... at least can reach equilibrium in a reasonable amount of time? – magnetar Nov 4 '16 at 4:48
• the smaller the cube, the quicker equilibrium should be reached, yeah. In general, we could as a first approximation solve the Fickian heat diffusion equation, which is a little bit of work though ... – Sanya Nov 4 '16 at 7:09