Environmental Science, CheriMathLab

🌎 Key Concepts & Laws

Carbon Cycle

Carbon flows between atmosphere ($\text{CO}_2$), biosphere (photosynthesis/respiration), oceans (dissolution), and lithosphere (fossil fuels, carbonates). Human activity has disrupted the balance.

Greenhouse Effect

$\text{CO}_2$, $\text{CH}_4$, and $\text{H}_2\text{O}$ absorb and re-emit infrared radiation. Radiative forcing: $$\Delta F = 5.35\,\ln\!\left(\frac{C}{C_0}\right)\;\text{W/m}^2$$ for $\text{CO}_2$.

Logistic Growth

$$\frac{dN}{dt} = rN\!\left(1 - \frac{N}{K}\right)$$ Population growth slows as it approaches carrying capacity $K$. S-shaped growth curve with inflection at $K/2$.

Ozone Chemistry

$\text{O}_3 + \text{UV} \rightarrow \text{O}_2 + \text{O}$. CFCs catalytically destroy ozone: $\text{Cl} + \text{O}_3 \rightarrow \text{ClO} + \text{O}_2$. One Cl atom can destroy ~100,000 $\text{O}_3$ molecules.

Biodiversity Indices

Shannon index: $$H = -\sum p_i \ln(p_i)$$ Measures species diversity accounting for both richness and evenness across ecological communities.

Ocean Acidification

$$\text{CO}_2 + \text{H}_2\text{O} \rightleftharpoons \text{H}_2\text{CO}_3 \rightleftharpoons \text{H}^+ + \text{HCO}_3^-$$ Increased $\text{CO}_2$ lowers ocean pH, dissolving carbonate shells and threatening marine ecosystems.

Energy Balance

Earth absorbs $\dfrac{S(1-a)}{4} = \sigma T^4$ where $S = 1361$ W/m$^2$ (solar constant), $a \approx 0.3$ (albedo). Equilibrium temperature ~255 K without greenhouse.

Renewable Energy Physics

Solar: $P = \eta \times A \times I$. Wind: $P = \tfrac{1}{2}\rho A v^3$. Betz limit caps wind turbine efficiency at 59.3%. Capacity factor measures real vs rated output.

✎ Problems

1 Easy

Calculate Earth's equilibrium temperature (no atmosphere) using the energy balance: absorbed solar radiation equals emitted thermal radiation. Use $S = 1361$ W/m$^2$, albedo $a = 0.30$, and $\sigma = 5.67 \times 10^{-8}$ W m$^{-2}$ K$^{-4}$. Compare to Earth's actual average temperature of $288$ K. What accounts for the difference?

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Absorbed $= S(1-a)\pi R^2$. Emitted $= \sigma T^4 \times 4\pi R^2$. Setting equal: $T = \left[\dfrac{S(1-a)}{4\sigma}\right]^{1/4}$. The ~33 K difference is the greenhouse effect.

2 Easy

A deer population in a forest has a carrying capacity of $500$. Starting with $50$ deer and an intrinsic growth rate $r = 0.15$ per year, use the logistic growth equation to find the population after 10 years. At what population size is the growth rate maximized?

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Logistic: $N(t) = \dfrac{K}{1 + \left(\dfrac{K - N_0}{N_0}\right)e^{-rt}}$. Maximum growth rate occurs at $N = K/2 = 250$. Plug in $t = 10$ years to find $N(10)$.

3 Medium

Atmospheric $\text{CO}_2$ has risen from $280$ ppm (pre-industrial) to $420$ ppm today. Using the radiative forcing formula $\Delta F = 5.35\,\ln(C/C_0)$ W/m$^2$, calculate the current radiative forcing from $\text{CO}_2$ alone. If climate sensitivity is $3°\text{C}$ per doubling of $\text{CO}_2$, estimate the equilibrium warming from this forcing.

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$\Delta F = 5.35 \times \ln(420/280) \approx 5.35 \times 0.405 \approx 2.17$ W/m$^2$. For a doubling ($\Delta F \approx 3.7$ W/m$^2$), warming $= 3°\text{C}$. So current warming $\approx 3 \times (2.17/3.7)\;°\text{C}$.

4 Medium

An ecosystem has five species with the following abundances: A = 40, B = 30, C = 15, D = 10, E = 5 (total = 100). Calculate the Shannon diversity index ($H$). Then remove species E and recalculate. How much diversity is lost?

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$H = -\sum p_i \ln(p_i)$ where $p_i = n_i/N$. For all 5 species: $H = -[0.4\ln(0.4) + 0.3\ln(0.3) + 0.15\ln(0.15) + 0.1\ln(0.1) + 0.05\ln(0.05)]$. Recalculate with 4 species (renormalize proportions).

5 Medium

A solar panel has an area of $2$ m$^2$ and efficiency of 20%. At a location receiving average solar irradiance of $5$ kWh/m$^2$/day, how much energy does it produce per day? Per year? If a household uses $30$ kWh/day, how many panels are needed?

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Daily energy $= \text{area} \times \text{irradiance} \times \text{efficiency} = 2 \times 5 \times 0.20 = 2$ kWh/day per panel. Panels needed $= 30/2 = 15$. Annual $= \text{daily} \times 365$.

6 Medium

Estimation Problem: The ocean absorbs about 30% of anthropogenic $\text{CO}_2$ emissions (currently ~40 Gt $\text{CO}_2$/year). The average ocean pH has dropped from $8.18$ to $8.07$ since pre-industrial times. If $\text{pH} = -\log[\text{H}^+]$, by what percentage have hydrogen ion concentrations increased? What does this mean for marine organisms with calcium carbonate shells?

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$\Delta\text{pH} = 0.11$. Since pH is logarithmic: $[\text{H}^+]$ ratio $= 10^{0.11} \approx 1.29$, meaning a 29% increase in hydrogen ion concentration. Higher $[\text{H}^+]$ shifts the carbonate equilibrium, making it harder for organisms to build $\text{CaCO}_3$ shells (lower saturation state $\Omega$).

7 Hard

A wind turbine has blade length $50$ m and operates at the Betz limit ($C_p = 16/27$). At wind speed $12$ m/s and air density $1.225$ kg/m$^3$, calculate the maximum power output. If wind speed drops to $8$ m/s, by what factor does power decrease? This illustrates why wind site selection is critical.

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$P = \tfrac{1}{2} \times C_p \times \rho \times A \times v^3$ where $A = \pi r^2 = \pi(50)^2$. Power scales as $v^3$, so the ratio $= (8/12)^3 = (2/3)^3 = 8/27 \approx 0.296$, power drops by ~70%!

8 Hard

Two competing species follow the Lotka-Volterra competition model. Species 1: $\dfrac{dN_1}{dt} = r_1 N_1\!\left(1 - \dfrac{N_1 + \alpha_{12}N_2}{K_1}\right)$. With $r_1 = 0.5$, $K_1 = 1000$, $r_2 = 0.3$, $K_2 = 800$, $\alpha_{12} = 0.6$, $\alpha_{21} = 0.8$. Determine the outcome: coexistence, competitive exclusion, or unstable equilibrium? Find the equilibrium populations.

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Coexistence requires $K_1/\alpha_{12} > K_2$ AND $K_2/\alpha_{21} > K_1$. Check: $K_1/\alpha_{12} = 1000/0.6 = 1667 > 800$? Yes. $K_2/\alpha_{21} = 800/0.8 = 1000 \geq K_1$? Yes (equal). At equilibrium, solve: $N_1 + 0.6N_2 = 1000$ and $N_2 + 0.8N_1 = 800$ simultaneously.

9 Hard

Carbon Budget: To limit warming to $1.5°\text{C}$, the remaining carbon budget is approximately $400$ Gt $\text{CO}_2$ (as of 2023). Current annual emissions are ~$40$ Gt $\text{CO}_2$/year. If emissions decrease linearly to zero, in how many years must we reach net-zero? If emissions instead decrease exponentially at 7% per year, do we stay within the budget?

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Linear: total $= \tfrac{1}{2} \times \text{base} \times \text{time} = \tfrac{1}{2} \times 40 \times t = 400$, so $t = 20$ years. Exponential: total $= \displaystyle\int_0^{\infty} 40\,e^{-0.07t}\,dt = \dfrac{40}{0.07} \approx 571$ Gt. This exceeds 400 Gt, so 7% annual reduction is insufficient for $1.5°\text{C}$.

10 Hard

The ozone layer absorbs UV-B radiation (280-315 nm). A single chlorine atom from a CFC molecule can destroy up to $100{,}000$ ozone molecules before being deactivated. If $1$ kg of CFC-12 ($\text{CCl}_2\text{F}_2$, $M = 121$ g/mol) is released, how many ozone molecules could potentially be destroyed? Express your answer in moles and molecules.

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$1$ kg CFC-12 $= 1000/121 \approx 8.26$ mol CFC. Each molecule has 2 Cl atoms: $16.53$ mol Cl. Each Cl destroys $10^5$ $\text{O}_3$: total $= 16.53 \times 10^5$ mol $= 1.653 \times 10^6$ mol $\text{O}_3 \approx 10^{32}$ molecules of $\text{O}_3$. This is why CFCs are so destructive.

11 Advanced

Ecological Modeling: A predator-prey system follows the Lotka-Volterra equations: $\dfrac{dH}{dt} = aH - bHP$ and $\dfrac{dP}{dt} = cbHP - dP$, where $H$ = prey (hares), $P$ = predator (lynx), $a = 0.4$, $b = 0.01$, $c = 0.5$, $d = 0.3$. Find the equilibrium populations. Describe the behavior near equilibrium (oscillations, stability). What is the period of population oscillations?

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Equilibrium: $H^* = \dfrac{d}{cb} = \dfrac{0.3}{0.5 \times 0.01} = 60$ hares. $P^* = \dfrac{a}{b} = \dfrac{0.4}{0.01} = 40$ lynx. The system oscillates with period $T \approx \dfrac{2\pi}{\sqrt{ad}} = \dfrac{2\pi}{\sqrt{0.12}} \approx 18.1$ time units. Lotka-Volterra gives neutrally stable cycles.

12 Advanced

Systems Thinking: Consider a simplified Earth energy balance model with ice-albedo feedback: $T = \left[\dfrac{S(1 - a(T))}{4\sigma}\right]^{1/4}$, where albedo depends on temperature, $a = 0.7$ for $T < 250$ K (snowball), $a = 0.3$ for $T > 280$ K (ice-free), and linearly interpolated between. Show that this system can have multiple stable equilibria. What are the implications for abrupt climate change?

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Plot the outgoing radiation ($\sigma T^4$) and absorbed solar radiation ($S(1-a(T))/4$) as functions of $T$. The intersections give equilibria. With temperature-dependent albedo, you get three crossings: a cold "snowball" state, an unstable middle state, and a warm state. The unstable middle equilibrium acts as a tipping point, small perturbations can push the system to a completely different climate state.