Sea Ice 101 – Insolated, Isolated, Insulated Icebergs in Space, A Mid-Term Exam

(Or, How Climate Science Is Taught Today)

Guest essay by Robert A. Cook

The following begins a series of topics on Arctic and Antarctic Sea Ice. And, since we (behind the keyboard) will be talking to you (the readers, and perhaps even the learners) as if we were openly addressing you in a classroom or hall, we will obviously begin class with the mid-term exam.

Don’t worry, the answers are already provided, but we are going to start the mid-term exam by questioning both the answers to the mid-term exam, but also the questions on the mid-term as they are already written.

More appropriately, we will question the ASSUMPTIONS and APPROXIMATIONS needed to make the “answers” work as written. Finally, we will assign the Sea Ice 101 Course Final Exam.

The class may now open its exam; any notes, articles and journals desired; the web, any available textbook or reference, and their minds. This particular mid-term exam is from the Institute for Structure and Nuclear Astrophysics (ISNAP), Department of Physics, University of Notre Dame. “The Physics of Climate” examples used were taught in spring semester, 2011, by Dr Michael Wiescher. Since Dr Wiescher is currently teaching a 2015 version of this class just down the hall, students are reminded to be civil, be quiet, and remain in their seats unless called upon. Guests may arrive in the classroom, they too are to be treated with all of the respect they earn.

Syllabus (2011) = http://isnap.nd.edu/Lectures/phys20054/class_syllabus.pdf

Lecture Notes (2015 class) = http://isnap.nd.edu/Lectures/phys20054/15Lecture%201%20Physics%20of%20Climate%20overview.pdf

20054 Mid Term Exam Solutions (2011) =

http://isnap.nd.edu/Lectures/phys20054/midterm_exam_solution.pdf

Lecture Notes, Earth Energy Budget and Balance (2011) =

http://isnap.nd.edu/Lectures/phys20054/Lecture_5_Energy_balance-1.pdf

Dr Wiescher’s lecture notes (Earth Energy Budget and Balance, hereafter EEBB 2011, are a fairly good summary and background of the earth’s heat budget. Incomplete, but a good start to some parts of the problem.

The mid-term based on those notes? A different story. In particular, let’s look at problem 4 on that exam, and the “solution set” for that problem. (I found this problem solution set in 2012, while looking for Arctic albedo calculations and assumptions.

The required solution to Problem 4 are reproduced below in their entirety.)

Let’s look at each error (or questionable assumption) in each sentence and phrase above.

1. “Consider the emission temperature of the polar ocean …”

1.A. Yes, “Emission temperature” is covered on page 15 of EEBB 2011. But the actual equations on pg15 do not include any of the formulas expected to be used when the student answers the question on the exam. Certainly misleading and confusing.

So, as we will see in a few minutes, even IF the student used this formula to answer Problem 4, that answer would be graded as “WRONG”!

1.B. But do the expected answers to Problem 4 actually calculate any “emission temperature”?

Regardless of the problem statement, the four required “answers” to Problem 4 are actually: Energy reflected _summer, energy reflected _winter, energy absorbed _summer, and energy absorbed _winter.

But look at pg 14 of the class notes (EEBB) for “energy absorbed” … (Not “emission temperature”)

We will see in a few minutes that even if the student used pg14 to calculate his or her answer to problem 4, that answer would still be graded as “WRONG” …. This is NOT the required answer either!

Finally, regardless of whether formulas from pg 14 or from pg 15 were used, an “emissions temperature” is still NOT a “heat flux”.

1.C. Next, regardless of what was “written” as the correct answer to Problem 4, or what was “expected to be written” as the answer to Problem 4, the ACTUAL information “IMPORTANT WHEN STUDYING CLIMATE CHANGE” when you answer Problem 4 is:

“4.A. What is the CHANGE in “Total Energy absorbed” AND “Total Energy lost” when sea ice is melted in the Arctic? Does your answer change between summer and winter conditions?”

Then, because the Antarctic sea ice behaves completely differently, Problem 4B is needed:

“4.B. What is the CHANGE in “Total Energy absorbed” AND “Total Energy lost” when sea ice is melted in the Antarctic? Does your answer change between summer and winter conditions? ”

2. “… for an incoming solar flux of F=1370W/m^2.”

2.A. First, the SORCE solar research team established a “new” correct solar irradiation level for Total Solar Irradiation (TSI) at the sun’s surface at 1362 Watt/m^2 in 2008. Here, the class is expected to use the wrong value for solar radiation levels.

The IPCC claims the entire “forcing” if atmospheric CO2 doubles is only 3.7 watts/m^2. Thus, by starting with a TSI radiation level “too high” by more than double the IPCC’s own 3.7 watts/m^2, every fundamental heat balance calculation after that statement is invalid and misleading. (Yes, 1370 watts/m^2 is the reference TSI used in the EEBB class notes, but that TSI value was already 3 years out-of-date when written.)

2.B. However, even IF the assigned 1370 watts/m^2 WERE the correct TSI radiation level, the actual radiation at top-of-atmosphere at the earth varies day-to-day over the year as the earth rotates around the sun in an elliptical orbit.

To calculate actual surface radiation levels at the earth’s surface, actual DAILY solar radiation levels at the top-of-atmosphere must be used.

Those ever-changing daily solar energy levels – which vary by 90 watt/m^2 from January through July – are NOT even expected to applied in the “approved answer” to Problem 4. But Problem 4 explicitly DOES require the student to compare heat balances at two different times of the year, so ANY season-to-season (really, day-by-day) change MUST BE correctly applied to the solar radiation at top-of-atmosphere. The earth does receive 1370 watts/m^2 at TOA twice a year: The first is March 23 (the spring equinox, near maximum Arctic sea ice extent each year) and 17 October, about a month after the fall equinox. But, the Arctic Ocean sea ice is in near total darkness (22 of 24 hours of the day) by that October 17.

If he or she chooses March 23, the sun is up in the sky above the Arctic, but the sun is NOT very high in the sky for very many hours each day, AND the Arctic sea ice is at its maximum extent for the year. Thus, ANY choice of a “summer conditions = smooth water” = “maximum solar energy absorbed” assumption for March 23 MUST be considered dead wrong.

So again, once again, IF the student answers Problem 4 accurately and completely, he or she will be graded “WRONG ANSWER” because neither March 23 nor October 17 is valid for the photographed conditions in Problem 4!

2.C. Atmosphere attenuation is ignored in the required answer to Problem 4.

In-atmosphere attenuation will reduce actual solar radiation levels at ground by MORE THAN 42% at noon in mid-September on the Arctic Ocean sea ice. (By 96.5% if noon, October 17 were chosen!) But atmospheric attenuation is ignored completely in the required answer to Problem 4. (Both air mass (length of passage) and atmospheric absorption (atmospheric clarity) need to be included when you calculate attenuation, and some assumptions have to be made for both, but the student MUST calculate something for attenuation loss before he can assume the solar energy hits the ice at sea level. )

So again, once again, IF the student answers Problem 4 accurately and completely, he or she will be graded “WRONG ANSWER.”

2.D. But the ocean surface is horizontal, right? And, any sea ice floating on a flat ocean surface is also horizontal.

So, to calculate “received radiation on a horizontal surface near the pole”, the student MUST correct for the altitude of the surface, the latitude of the surface, and the earth’s axial tilt on the day the surface is struck by sunlight. Now, these are not “original” discoveries to science – the formulas have been in use even before Magellan’s crews circled the globe, but they have to be done. (And be done properly. One problem: Even the 20054 class notes cover this topic very poorly, so the 20054 class notes need to be addressed as well.)

Yet again, IF the student spends the extra time to answer Problem 4 accurately and completely, he or she will be graded “WRONG ANSWER.”

3. “Choose the albedo and emissivity parameters for summer and winter conditions….”

3.A. Albedo changes over the year for Arctic sea ice are shown on pages 3,4,5, and 8 of EEBB 2011. In fact, pages 4 AND 8 are actually taken from Perovich 2002 (see also Curry 2001) during actual SHEBA albedo measurements on the actual arctic icecap. Clearly, the “right answer” for Arctic sea ice albedo over the summer arctic ocean is available to the class. (We do not know if it was introduced accurately, or whether the in-class discussion was accurate and detailed, but it was available to the class. Has been discussed in class. )

But the required “answer” for Problem 4 ignores BOTH the EEBB sea ice albedo plot, AND the sea ice photographs used in class.

Instead, the “answer” to Problem 4 uses a single “average” albedo for “ice” (obviously assuming “winter = ice”) of 0.80. Worse, as shown in few minutes, even this expected “winter albedo” of “sea ice” in Problem 4 is itself wrong!

3.B. Still worse than those two problems, getting the correct “answer” to Problem 4 requires assigning an albedo of 0.06 for “summer conditions”, again requiring the student to equate “summer (in the arctic) = only open ocean water (in the Arctic).”

Now remember your geography. All of the Arctic Ocean is above the Arctic Circle. So, from the middle of October through the end of February (the “winter conditions” for problem 4) there is NO sunlight at all ANYWHERE across the Arctic Ocean sea ice. It is completely dark all day that far north. Therefore, the photograph in Problem 4 showing “winter conditions” was actually taken sometime during the arctic summer, and – yes! – it’s cloudy, wind-swept snow and block ice does show what 80% of the arctic “summer” days are like on the icecap !

So, yet again, if a student answers Problem 4 accurately and to the best of his or her ability, he or she gets graded with a “WRONG ANSWER”.

3.C. Worse, 0.80 is not the correct value for winter sea ice, mid-summer sea ice, snow-covered sea ice, Arctic sea ice nor Antarctic sea ice under any conditions or seasons.

Again, look at the EEBB plot of measured Arctic sea ice albedo on page 4: The summer “ice” albedo varies by date: beginning as high as 0.85 in May, going as low as 0.38 to 0.40 in mid-July 30 through August 12, returning to 0.75 in late August, then rising back towards 0.85.

Using this plot from their class notes, even a high school student “should” be able to pick a date, then chose the correct “sea ice” albedo for that date and that latitude, right? These are college students, paying $45,000.00 (or more) per semester to learn the “Physics” or “Climate Science”. If I pay a beginning engineer $65,000.00 yearly salary to calculate solutions in the real world, a student paying 90,000.00 per year should be required to read a single graph in a single class.

Therefore, the correct summer and winter sea ice albedo should be required to answer Problem 4, right? Well, the “summer conditions” albedo is not correct.

3.D. Well, 0.07 is not the correct albedo for water either.

The albedo of the open ocean varies from 0.06 up to 0.45 (and higher) by:

solar elevation angle (or solar zenith angle);

ratio of direct and diffuse solar radiation present;

and by the wind speed and area of water present (which controls wave height).

Specifically BECAUSE the Arctic sunlight levels change so much from winter to spring to summer to fall and back to winter in both elevation angle AND duration (how high the sun any given day and how long the sun is visible each day), the student MUST be able to calculate solar elevation angles for ANY day of the year at ANY time of day.

In Problem 4, the need is particularly important: The albedo of open ocean water changes from 0.06 (at noon, in mid-June or July) to 0.35 – 0.40 (each morning, evening and night in the summer; all day in the spring and fall). Once again, IF the student answers Problem 4 accurately and completely, he or she will be graded “WRONG ANSWER.”

4. “…. as shown in the pictures …”

4.A. We will repeat a little bit from above. Let’s look at the pictures in more detail.

The photograph for “summer conditions” shows absolutely calm water( no waves, no winds), no clouds (direct radiation only), and clear skies (no aerosols or particulates implied.) The near-absence of shadows even on the highest icebergs imply at least a 36 – 45 degree solar incidence angle. These conditions mandate the student assign a different albedo for BOTH “summer ice” and “summer ocean” conditions!

4.B. The presence of several high icebergs in the background of the “summer conditions” means that photograph was taken much further south than the “arctic Ocean.” (The clumpy, broken large chunks of ice floating in the foreground of the same image show the same evidence, but not as strikingly as the three big icebergs.) This is because large stable icebergs do not form in the central arctic. Large icebergs actually break off from mountain glaciers found down both coasts of Greenland, then drift south with the ocean currents in the Davis Strait and Denmark Strait.

ALL sea ice in the Arctic itself (where the “arctic sea ice” loss is measured!) forms as low 1 and 2 meter thick ice. Some 25-35% of first and second year sea ice is 2 meters thick, this sea ice remains through the summer and forms flat-bottomed 3, 4 and 5 year sea ice. This multi-year sea ice is often trapped against the northern Canadian coasts and islands by winds and currents. At times, the flat “sea ice” is pressed together into thin, tall ridges, but when these ridges are freed and allowed to float freely, they immediately fall sideways back to “flat sea ice”. When the Arctic sea ice flows south, these pieces stay flat. Thus, there are no icebergs in the Arctic Ocean – they simply don’t form up there. Arctic sea ice is flat 2-3 meter floes, with less than 30 cm of ice above water.

The presence of icebergs, calm seas (and the ocean is notoriously rough east of Greenland, in the Bering Sea, and north of Iceland) and a high solar elevation angle indicate that this photograph was taken in mid-summer about latitude 50-60 north – probably near the Labrador sea.

And the Labrador Sea, the St Lawrence River estuary, the Davis Straits are never ice-covered in summer, and some areas of each just ” touched” by sea ice in much of winter.

So, no student should compare “summertime” heat absorbed in an open ocean under clear skies near Labrador with an Arctic Ocean sitting in the dark under ice 20 degrees further north! Once again, IF the student answers Problem 4 accurately and completely, he or she will be graded “WRONG ANSWER.” (It would be like comparing the sun tan a student gets in Miami Feb 10 with one in Indiana the same day.)

4.C. The “winter conditions” photograph is even more mysterious. The Arctic sea ice between mid-October and mid-March is far north of the Arctic Circle, and thus has NO sunlight exposure at all at ANY time of day. The photo shows high winds (blowing ice in ½ to 1 meter swirls), a light gray even diffuse light (as if there were no heavy storm clouds at all, despite the evidence of high winds!), and a rough mid-ocean type of low pressure ice ridges.

4.D. The shadows in this picture are lightly shaded coming from the low right at perhaps 15-20 degrees. This implies either a early April “noon” sunlight at the time of maximum Arctic sea ice area, or a limited area further south such as Hudson Bay (near latitude 60 north) sometime between mid-March and mid-May. Hudson Bay completely melts by Mid-July, so it is not likely to be a late May or June timeframe.

4.E. The ice covering in the “winter conditions” means there is no evaporation losses, no convection losses from the open water, and increased conduction temperature differences between the ocean water under the ice and the atmosphere. Further, the presence of high-level clouds means significantly MORE diffuse solar radiation – which changes both the albedo of the sea ice and that of the open water the sea ice supposedly replaces. But higher wind speed increases convection losses over the “summer” calm ocean waters, but would increase evaporation losses if water were present. (Conflicting, isn’t it?)

Worse, the presence of clouds is MORE likely over open ocean waters (as in the summer conditions photo) and less likely over ice-covered seas (the winter conditions photo). Again, the photos contradict hundreds of days of actual on-site observations of both Arctic and Antarctic winter and summer conditions! (See the lecture notes on EEBB pg 4 (from Perovich, 2002) for photographs of actual winter (pre-summer melt) and summer sea ice conditions. Clearly, the class has been shown photo’s of what real sea ice looks like under real summer conditions. Changing fundamental information in the middle of a test problem is hardly reasonable nor prudent if you want new students to understand a complex subject.)

Cloud albedo also affects solar radiation available at ground level, and the effect of cloud albedo differences is covered in the class notes, EEBB pg 5. But cloud albedos are ignored here in the Problem 4 solution set, where clouds matter a great deal in all parts of the problem solution!

Once again, IF the student answers Problem 4 accurately and completely, he or she will be graded “WRONG ANSWER.”

4.F. The actual daily AVERAGE air temperatures vary from month-to-month in both the Arctic and Antarctic. Hourly temperature changes – even at the same daily “average” temperatures! – are much greater under clear skies with no clouds and low winds than under cloudy skies with high winds. Dry bulb temperature, wet bulb temperature, wind speed, relative humidity and air pressure all change relative humidity calculations. Relative humidity and wind speed changes greatly affect the film temperature and heat losses from both ice-covered surfaces AND from the open ocean in both winter and summer conditions.

But NO convection, conduction, evaporation, nor radiation heat losses or gains are addressed in the problem 4 solution set.

Once again, IF the student answers Problem 4 accurately and completely, he or she will be graded “WRONG ANSWER.”

4.G. Long wave LW radiation losses again differ significantly in both picture conditions – again in opposite directions. Relative humidity and surface skin temperature and sky temperatures (no clouds, high thin clouds, or low dark clouds) ALL change how much long wave radiation is lost to space each hour.

Again, these real-world differences and their effect on the radiation losses are NOT covered in the required “answer set” to this problem. IF the desired answer was “emission temperature” under both conditions, or temperatures of any kind under either conditions, then all heat exchange MUST be addressed and correctly calculated for full credit to the problem.

More fairly, such calculations are a bit long, so a full calculation should either be given more time, or a more limited question asked that can be answered in the time provided.

Once again, IF the student answers Problem 4 accurately and completely, he or she will be graded “WRONG ANSWER.”

5. Finally, having probably spent entirely too much time looking at just the problems within Problem 4 in the problem statement of Problem 4, let’s look at the “equation solutions” of problem 4. First, look at how the student is expected to calculate the reflected energy in the summer and in winter.

Summer albedo ~ 0.07 F_ref = 0.07 * 1370 ~ 96 W/m^2

Winter albedo ~ 0.80 F_ref = 0.80 * 1370 ~ 1096 W/m^2

My judgment? If you answered both parts as required by the answer sheet, you should flunk. But, IF the student answers Problem 4 accurately and completely, he or she will be graded “WRONG ANSWER.”

5.A. Both albedos are incorrect.

First, 0.07 is not the correct albedo for water. 0.06 is closer, but even that value is ONLY valid for wind-swept sea water under diffuse radiation, NOT the very calm summer arctic ocean water in direct sunlight pictured

The albedo for “pure water” at various solar elevation angles (not even solar zenith angles) is “somewhat” and very poorly shown in EEBB pg 6 below. (Various albedo changes for several different substances at difference solar zenith angles are also shown on class handout EEBB pg 7, but, admittedly, these plots are for solid land varieties.)

However, even this attempt at displaying Fresnel’s graphs is misleading: The plot shown is valid. But ONLY for Fresnel’s approximations of two perpendicular polarized light waves showing the theoretical reflections from a perfectly calm pure water surface at room temperature in the lab. The real-world albedo for the actual “open ocean” has been measured at sea, in the arctic leads, and from open water towers since Payne published his first papers in 1972. The values for open ocean albedo are referenced by Perovich (and Curry and many other sources) in the documents already available to the student. There is no excuse for allowing anybody anywhere to even imply Fresnel equations are valid for ocean waters.

The actual albedo of calm ocean water in mid-September (80 degree north, at noon, when the arctic sea ice is at its minimum and the sun is at its highest point) is 0.316.

So, the expected albedo for “summer conditions” in problem 4 is wrong. Is the expected albedo for “winter conditions” in Problem 4 any better?

Well, no.

0.80 is NOT accepted for ANY period of the year in either hemisphere for sea ice under ANY conditions.

0.83 is the measured albedo for both Arctic and Antarctic winter sea ice covered with fresh snow – but only until late spring in either hemisphere. Arctic sea ice albedo drops down to 0.40 in mid-summer. Bare sea ice (if snow is blown off by high winds) is about 0.60; but bare sea ice rarely occurs at any time of year.)

Antarctic sea ice albedo also goes down during the melt season, but not as far. Its albedo drops to 0.75 in mid-summer (Dec-Jan-Feb). Again, look at the plots in the class notes, pg 4. [See Brandt and Warren, 2005; Curry, 2001, Perovich, 2002 for sea ice albedos; and Burt, 1954; Budyko, 1956; Grishchenko, 1959; Davies, 1969; Payne, 1972; Perovich, 1984; Briegleb, 1986; Frank Hansen, 1993; Bacellar, 2002; Rutlege & Schuster, 2004; Jin, 2004; Pegau & Paulsen, 2008 (and others) for open ocean albedos. ]

5.B. The TOA solar radiation level in Problem 4 is wrong. (Since TOA = 1370 watt/m^2, the solar radiation levels are for TOA levels in mid March or mid-October ONLY.)

5.C. No atmospheric attenuation nor air mass calculation is made. Thus, the student is expected to assume that the sea ice is being irradiated up in space, right?

5.D. No latitude correction is made. Thus, the student must assume that the ocean water is tilted perpendicular to the sunlight, right?

Oddly, Problem 7 on the same mid-term exam does require the student to calculate atmospheric attenuation – although the method required in Problem 7 is not valid for other latitudes and air clarity values – but clearly the students are expected to know how to make the calculation correctly. And, Problem 3 on the mid-term exam requires the student to address albedo differences on different days of the year. (Perhaps they are just not required to know WHEN to make to make the correct calculation the correct way using the correct values.)

6. F_absorbed.

Frankly, this calculation puzzles me. Clearly, the intent is to apply Kirchhoff’s Law to the absorbed energy going into the sea ice as shown in the class notes, EEBB, pg 2.

But the assumptions required when you derive Kirchhoff’s law mean it is ONLY valid for a perfect blackbody perfectly radiated in a perfect black hole in a perfect space.

The only input energy allowed in is from solar radiation at a perfect temperature spectrum and uniform absorptivity, perfectly emitted from a radiating black body onto a perfect “grey body” of a perfectly uniform emissivity and perfectly uniform heat capacity. (Thus, the sea ice must be in space, directly exposed to the sun’s “average” energy.)

The area receiving solar (SW or IR) radiation must be the same size and orientation as that radiating thermal (LW) radiation. (No heat loss is permitted from the back wall of the sea ice. No radiant losses in any other direction are permitted when Kirchhoff’s Law is derived as shown.)

Both receiving and radiating areas must be exactly perpendicular to the original radiating blackbody. (No correction is made for angle of the sea ice to the radiant energy coming in. radiant energy outbound is assumed perpendicular to the gray body. )

The “viewing angle” between both black bodies is perfect (No occluded or hidden surfaces. This goes with the perpendicular assumption above, and is hard to accomplish in real-world spacecraft radiating energy from their power sources.)

The perfect gray body radiating energy must be at perfect thermal equilibrium with the space around it. (In this particular derivation, the surrounding space is itself a perfect black body at 0.0 K degrees. All parts of the gray body are assumed at the same temperature.)

No other heat loss can be allowed when you use Kirchhoff’s Law. (Hence, my claim that the iceberg is “insulated” and “isolated” on the front, back side and all walls. No evaporation, no convection, no conduction losses permitted. No heat loss to the atmosphere, to the salt water underneath, no sublimation, no phase changes allowed in melt ponds above the surface, no freezing of sea water below the surface if air temperature lowers. Completely isolated sea water.)

No light energy is allowed to go through the sea ice. (Opacity is 0.0. This condition, at least, is somewhat reasonable.)

But the required equation itself, after the student is shown Kirchhoff’s Law?

Energy absorbed = (1-albedo) * (radiation available at TOA) * (1-emmisivity)

No. Absolutely not.

But a good equation for opaque, insolated, isolated, insulated icebergs in space however.

7. Final exam, Sea Ice 101.

There will be a one-question Final Exam when the Sea Ice 101 Class ends later this year. Students should keep their notes ready for this question.

Calculate the actual energy absorbed and energy released from the open water and from sea ice on ONE date and ONE latitude at any selected single hour. Chose both Arctic or Antarctic latitude, and any day-of-year – BUT! – sea ice must actually be exposed to solar radiation on that hour of that day! (You may not select 85 north on Dec 22. You may not select 85 south on 22 Dec. You may NOT select 0.0 latitude on any day-of-year.)

Further, since all Final Exam answers will be available to all members, you cannot select any latitude or day-of-year already chosen by somebody else. (First chosen, first choice.)