ERTH 535: Planetary Climate Change (Spring 2018) Class Meeting Summary Mon. & Fri., Feb. 19 & 23 Dr. Dave Dempsey Dept. of Earth & Climate Sci., SFSU

1. Energy and Transformations of Energy

(See in-class PowerPoint presentation (provided here as a PDF file).

2. The Concept of an Energy Budget and a Simple Application to the Earth as a Whole (i.e., atmosphere and surface, including oceans, and the interior)

To understand the behavior of the temperature of any object, including the earth as a whole (where global average temperature is one aspect of climate), we must consider not only the single source of energy that we've considered so far (absorption of incoming solar radiation) but also other sources of energy and ways to lose energy simultaneously. That is, we must construct an energy budget. More specifically, we want to construct a budget for heat in the earth system as a whole (since temperature is related directly to heat content, not energy content considered more generally):

A generic heat budget equation looks like:

 Rate at which an object's heat content changes = Sum of rates at which an object gains heat by various mechanisms ("sources") — Sum of rates at which an object loses heat by various mechanisms ("sinks")

The same expression applies if the word "heat" is replaced by "mass" (e.g., amount of water in a water budget) or "momentum" (a momentum budget, which is all that Newton's Second Law is). Equations of this form are mathematical expressions of conservation laws. The word "conservation" here seems odd, given that the terms in the equation are all about rates of change. However, if the two terms on the right-hand side balance each other, then the net rate of change (the term on the left-hand side) will be zero—that is, the object's heat (or mass or momentum) content will be conserved (that is, not change with time). In that case, the object's heat (or mass or momentum) budget is balanced or at equilibrium or at steady state.

It's worth emphasizing (because the terms are easy to misinterpret) that in the context of a budget equation like this one, a "source" does not mean "a place from which heat comes" but rather a mechanism by which the object gains heat (which, to complete the description of the mechanism, might, or might not, include information about the place or object from which the heat or other energy comes). Similarly, a "sink" is not "a place where heat goes", but rather a mechanism by which the object loses heat.

It's also worth noting that sensible heat (the total kinetic energy of random motions of molecules in a substance) and temperature (a measure of the average kinetic energy of random molecular motions in a substance) are closely related, but of course not the same. The relationship is a proportional one:

 T (absolute temperature of an object) ∝ (i.e., is proportional to) H (heat content of the object)

"Proportional to" means that if you change the quantity on the right-hand side by some multiplicative factor—say, 2 or 0.3 or whatever—then the quantity on the left-hand side changes by the same factor. A proportionality relation such as this one can be converted into an equality by introducing a constant of proportionality on the right-hand side:

 T (absolute temperature of an object) = constant × H (heat content of the object)

Since heat is the total kinetic energy of random molecular motions in an object or material, which should depend on the amount of material (in particular, its mass), whereas temperature is the average kinetic energy of random molecular motions in the same object or material (which doesn't depend on the amount of material involved), it should make sense that the constant of proportionality should depend on the mass of the object or material. Although perhaps not so obvious, the constant also depends on the type of material (liquid water, air, ice, rock type, etc.). (The "constant" is not really quite constant, is it!) We'll see more about this relation between an object's absolute temperature and its heat content later in the course.

It's worth noting that this relationship implies that if an object's heat content changes, then so will its temperature. More generally, for an object's temperature to change, it it must gain or lose heat. We can extend this idea very simply to relate the rate at which an object's temperature change to the rate at which its heat content changes, and introduce that version of the relation into the generic heat budget equation above:

 Rate at which an object's temperature changes ∝ (i.e., is proportional to) Rate at which an object's heat content changes = Sum of rates at which an object gains heat by various mechanisms ("sources") — Sum of rates at which an object loses heat by various mechanisms ("sinks")

This relationship tells us that the rate at which an object's temperature changes is proportional to the difference between the rates at which the object gains and loses heat—the bigger the imbalance, the faster the object's temperature will change.

We can apply this general relationship to the earth as a whole (or to the atmosphere, oceans, and outer shell of the solid earth, which are the parts most involved in climate, except on very long time scales), to help us understand some important questions about how and why climate changes (at least, insofar as temperature goes). So far we have focused on only one source of heat for the earth—absorption of solar radiation. A more complete list of sources of energy for the earth, expressed as fluxes instead of rates (for example, see Skeptical Science, "Heat from the Earth's Interior Does Not Control Climate"), is:

1. absorption of solar radiation (341.3 W/m2);
2. radioactive decay in the solid earth (converting nuclear energy into heat) (~0.09 W/m2);
3. fission in nuclear power plants (also converting nuclear energy into heat) and combustion of fossil fuels by humans (converting chemical potential energy into heat) (~0.03 W/m2);
4. frictional dissipation associated with wind, ocean currents, tides, plate tectonics, and other internal motions (converting kinetic energy into heat) (~0.007 W/m2 from tides);
5. respiration of living organisms, including plants (~0.002 W/m2);
6. absorption of radiative energy from distant stars (extremely small);
7. etc.

As you can see from the figures above, for the earth as a whole solar absorption is a far greater source of energy than anything else, accounting for 99.97% of the total, so for the purpose of studying climate we often need to consider only the sun as a source of energy.

Sinks of heat for the earth constitute a much shorter list:

1. radiative emission (by the earth's surface and atmosphere) to space

Strictly speaking, we should include phase changes of water as sources and sinks of heat for the earth, because evaporation, melting, and sublimation all convert heat into latent heat, thereby constituting sinks of heat for the planet, and condensation, freezing, and deposition all convert latent heat into heat, thereby constituting sources of heat for the planet. These are certainly important for understanding aspects of the internal workings of climate. However, on a global average, these processes largely compensate for each other and the difference between them (the net impact on the heat budget) is relatively small. For a warming planet, phase changes of water constitute a net sink of heat because more water evaporates from the oceans than condenses to form clouds, and snow and ice melt from glaciers, ice caps, and sea ice faster than water freezes to re-form them. Conversely, when the earth enters cools and enters a glacial period (sometimes called an ice age), water transfers from the oceans to polar ice sheets, creating a net conversion of latent heat in the water to sensible heat in the atmosphere. But these transitions take thousands of years and are much too slow to show up as a significant source or sink of heat compared to solar radiation absorption and radiative emission. More generally, the difference between evaporation, melting, and sublimation on one hand (sinks of heat for the planet) and condensation, freezing, and deposition on the other hand (sources of heat for the planet) is small enough that we'll neglect them for now and focus the dominant contributors to the planet's heat budget, namely absorption of solar radiation and emission of radiative energy.

With the foregoing caveat, the only really significant way that the earth loses heat is by emitting radiative energy to space.

Over sufficiently long time (multiple years to decades or longer), energy gained by solar absorption very nearly balances the energy lost by radiative emission. Hence, in the global average, over a sufficiently long time:

 0 ≈ Average rate at which the earth absorbs solar radiative energy — Average rate at which the earth emits radiative energy to space

(where "≈" means "approximately equal to").

Note that the balance/equilibrium/steady state suggested here is not quite exact. If we want to estimate the rate of change of global average heat content of the earth, for this relationship to be useful we have to quantify the terms in it; otherwise we can't tell whether absorption exceeds emission or vice-versa and we can't tell in what sense the earth's heat content is changing (that is, we can't tell the sign of the left-hand side of the equation), much less how fast it might be changing (the magnitude of the left-hand side). However, we can't measure either term directly.

Instead, our strategy for quantifying the terms will have two steps: (1) relate each term to other quantities that we can measure or determine by other means; and (2) quantify those quantities to calculate the terms in the budget equation above. For example, the derivation of the inverse square law below is a step toward quantifying the average rate at which the earth absorbs solar radiative energy.

3.

4. Good Problem-Solving Strategy (separate handout)
1. Example of a Physical Relation: Inverse Square Law for Solar Radiative Energy Flux

(See handout: Inverse Square Law [PDF file].)

This is a version of the inverse square law for solar radiative energy flux. It says that the radiative energy flux from a point (or near-point) source (such as the sun) varies inversely with the square of the distance from the center of the source of the energy.

2. Sample Problem: Solar Radiative Energy Flux (Insolation) at the Earth's Average Distance from the Sun

[This problem applies the format described by the "Good Problem Solving Strategy" handout and that Problem #1 asks that you use.]

The sun's radius is 696,000 km and its surface temperature is about 5780 K. The average distance of the earth from the sun is 149,600,000 km. What is the flux of solar radiative energy at this distance from the sun at the top of the earth's atmosphere, directly facing the sun, at the earth's average distance from the sun?

I. Info Given or Otherwise Known:

• Radius of the sun (or the distance from the center of the sun to its surface) ≡ rsun = 696,000 km = 6.96 x 105 km
• Temperature of sun's surface ≡ Tsun= 5780 K = 5.78 x 103 K
• Average distance of earth from sun ≡ re2s(avg) = 149,6000,000 km = 1.496 x 108 km
• σ ≡ Stefan-Boltzmann constant = 5.67 x 10-8 W/(m2K4)

II. Info Desired:

• Intensity (flux) of solar radiation on a surface facing the sun at the average distance of earth from the sun ≡ S(re2s(avg))

III. Relations Needed:

1. S(re2s(avg)) = S(rsun) x (rsun/re2s(avg))2
(Inverse square law for solar radiative energy flux, applied to earth's average distance from the sun, where S(rsun) is the radiative emission flux from the sun's surface)

2. S(rsun) = σTsun4
(Stefan-Boltzmann Law, applied to the sun's surface, where Tsun is the absolute temperature of the sun's surface)

IV. Solution:

1. Substitute relation Eq. (2) into Eq. (1) to eliminate S(rsun):
 S(re2s(avg)) = σTsun4 (rsun/re2s(avg))2

[This is the symbolic solution because the information desired is on the left-hand side and we know the value of every quantity appearing on the right-hand side.]

2. Substitute numerical values and units:

S(re2s(avg)) = 5.67 × 10-8 W/(m2K4) × (5.78 x 103 K)4 × [(6.96 × 105 km)/(1.496 × 108 km)]2

3. Separately group numbers, powers of ten, and units; simplify units; simplify powers of ten; and lastly, compute a numerical answer:

 S(re2s(avg)) = [5.67 × 5.784× (6.96/1.496)2] × [10-8 × (103)4 × (105/108)2] [((W/m2)/K4)×K4×(km/km)2] = [5.67 × 5.784 × (6.96/1.496)2] × [10-8 × 1012 × 10-6] W/m2 = [5.67 × 5.784 × (6.96/1.496)2] × 10-2 W/m2 = 136978 × 10-2 W/m2
 S(re2s(avg)) = 1370 W/m2

V. Check Solution:

Units are those of energy flux; the value is positive, as we'd expect; and the value agrees closely with the value given by the text, which is based on satellite observations. It is reassuring that this theoretical calculation (based on the sun's surface temperature) agrees so closely with the observations. This supports (but doesn't prove) the validity of the theoretical calculation.

3.

4. Another sample problem: Problem #2(b) on p. 56, Chapter 3 of our text (The Earth System, 3rd Ed.)

There are two approaches to this problem. First, since the text derives the relation that is needed to solve this problem, you can represent a solution to Problem #2(a) symbolically, substitute into the text's symbolic solution for 2(b), and substitute values and units into the resulting solution to calculate the requested values.

Another approach is to start farther back and show formally how the text got the relation that it presents. I prefer doing this, because it shows how the solution is based on more fundamental physical and geometric relations.

I. Info Given or Otherwise Known:

• Distance between Mars and the sun ≡ rM→s = 1.52 AU (astronomical units)
• Distance between Venus and the sun ≡ rV→s = 0.72 AU
• Albedo of Mars ≡ αM = 0.22
• Albedo of Venus ≡ αV = 0.8
• Mean distance between the earth and the sun ≡ re→s(avg) = 1.0 AU
• Solar radiative flux at the top of the earth's atmosphere facing the sun at the earth's mean distance from the sun ≡ S(re→s(avg)) = 1370 W/m2
• σ ≡ Stefan-Boltzmann constant = 5.67 x 10-8 W/(m2K4)

II. Info Desired:

1. Effective radiating temperature of Mars ≡ TM
2. Effective radiating temperature of Venus ≡ TV

III. Relations Needed:

1. 0 ≈ SRabs – ER
(Approximately balanced heat budget equation for a planet)

where:

SRabs ≡ a planet's rate of absorption of solar radiation, and
ER ≡ the rate at which a planet emits radiative energy

2. Rate = Flux × A
(General relation between a rate and a flux, where A ≡ surface area)

3.

4. Sabs = Sarr × (1 – α)
(Relation between fluxes of solar radiation absorbed by and arriving at a surface)

where:
Sabs ≡ the flux of solar radiation absorbed by a surface
Sarr≡ the flux of solar radiation arriving on a surface
α ≡ the albedo of the surface

5. S(r) = S(rref)×(rref/r)2
(Inverse square law for solar radiation flux)

where:
r ≡ distance from the sun
S(r) ≡ solar radiation flux (normal to the sun's rays) at distance r from the sun
rref ≡ (known) reference distance from the sun
S(rref) ≡ (known) solar radiation flux (normal to the sun's rays) at distance rref from the sun

6. EF = σT4
(Stefan-Boltzmann Law)

where:
EF ≡ radiative emission flux from a blackbody
T ≡ absolute temperature of the blackbody

7. AX-sect = πR2
(Geometric relation between the surface area and radius of a circle)

where:
AX-sect≡ area of a circle (or cross-sectional area through center of a sphere)
R ≡ radius of the circle

8. Asfc = 4πR2 (Geometric relation between the surface area and radius of a sphere)

where:
Asfc≡ surface area of a sphere, and
R ≡ radius of the sphere

IV. Solution:

1. Apply Eq.(2) to SRabs and ER, respectively, in Eq. (1):

1. 0 ≈ Sabs× AX-sect – EF × Asfc

where:
AX-sect ≡ cross-sectional area of a planet (which is a circle, the effective shape that a planet presents to the sun as viewed from the sun)
Asfc ≡ surface area of a planet
EF ≡ radiative emission flux from the planet
Sabs ≡ solar absorption flux on a surface facing the sun directly

2. Apply Eq.(3) to the first term in Eq. (8) and Eq.(5) to the second term:

1. 0 ≈ S(rplanet→s) × (1 – αplanet) × AX-sect – σ(Tplanet4) × Asfc

where:
rplanet→s ≡ the distance from the planet to the sun
S(rplanet→s) ≡ the solar radiation flux at the top of the planet's atmosphere, facing the sun
αplanet ≡ the albedo of the planet
Tplanet ≡ the effective radiating temperature of the planet

3. Apply Eq.(4) to the first term in Eq.(9), using the earth's mean distance from the sun and the earth's solar constant as the reference solar radiative flux:

1. 0 ≈ S(re→s(avg))×(re→s(avg)/rplanet→s)2 × (1 – α) × AX-sect – σ(Tplanet4) × Asfc

4. Apply Eq.(6) to the first term in Eq. (9) and Eq.(7) to the second term:

1. 0 ≈ S(re→s(avg))×(re→s(avg)/rplanet→s)2× (1 – α) × π(Rplanet2) – σ(Tplanet4) × 4π(Rplanet2)

where Rplanet is the radius of the planet.

5.

6. Divide Eq.(10) by π(Rplanet2):

1. 0 ≈ S(re→s(avg)) × (re→s(avg)/rplanet→s)2 × (1 – α) – σ(Tplanet4) × 4

7. Solve Eq.(12) for Tplanet4:

1. Tplanet4 ≈ S(re→s(avg)) × (re→s(avg)/rplanet→s)2 × (1 – α) ÷ (4σ)

8. Solve Eq.(13) for Tplanet by taking the fourth root of the equation:

1. Tplanet ≈ [S(re→s(avg)) × (re→s(avg)/rplanet→s)2 × (1 – α) ÷ (4σ)]1/4

9. Apply Eq.(14) to Mars and Venus:

1.  TM ≈ [S(re→s(avg)) × (re→s(avg)/rM→s)2 × (1 – αM) ÷ (4σ)]1/4

2.  TV ≈ [S(re→s(avg)) × (re→s(avg)/rV→s)2 × (1 – αV) ÷ (4σ)]1/4

(Note that these are versions of the relation derived by the text in Chapter 3, but now you should have a better idea of where it came from!)

10. Substitute values and units into Eqs.(15) and (16):

(and so on from here)

V. Check solution:

(Check units, sign, and magnitude of the solution. The units should be in Kelvins and the sign should be positive (because the absolute temperature cannot be a negative number). The magnitudes should be physically possible (neither extremely cold nor extremely hot).