Editing Openai/692b0130-6a64-800a-8103-b4de485f6aa0 (section)

=== How I calculated it (step-by-step arithmetic) ===

Assumptions
* World population ≈ 8.0×10⁹ people.
* Each person pees 0.5 L = 0.0005 m³ in that one event (reasonable single-event volume; results scale linearly).
* Lake Superior volume ≈ 12,100 km³ = 12,100 × 10⁹ m³ = 1.21×10¹³ m³ (standard approximate value).
* Density of water = 1000 kg/m³.
* Urine (initial) temperature ≈ 37.0 °C; lake temperature assumed 4.0 °C (so ΔT = 33.0 K).
* No heat losses to air/ground; perfect mixing (this is a conservative simple energy-balance model).

Step math
# Total urine volume people × urine per person = 8.0×109×0.0005 m3=4.0×106 m38.0\times10^{9} \times 0.0005\ \text{m}^3 = 4.0\times10^{6}\ \text{m}^38.0×109×0.0005 m3=4.0×106 m3.
# Masses (density 1000 kg/m³) - Mass of urine = 4.0×106 m3×1000 kg/m3=4.0×109 kg4.0\times10^{6}\ \text{m}^3 \times 1000\ \text{kg/m}^3 = 4.0\times10^{9}\ \text{kg}4.0×106 m3×1000 kg/m3=4.0×109 kg. - Mass of lake = 1.21×1013 m3×1000 kg/m3=1.21×1016 kg1.21\times10^{13}\ \text{m}^3 \times 1000\ \text{kg/m}^3 = 1.21\times10^{16}\ \text{kg}1.21×1013 m3×1000 kg/m3=1.21×1016 kg.
# Final temperature (mixing) — using conservation of energy: Tfinal=mlakeTlake+murineTurinemlake+murineT_{\text{final}} = \dfrac{m_{\text{lake}}T_{\text{lake}} + m_{\text{urine}}T_{\text{urine}}}{m_{\text{lake}}+m_{\text{urine}}}Tfinal=mlake+murinemlakeTlake+murineTurine. Temperature increase ≈ murinemlake+murine×(Turine−Tlake)\dfrac{m_{\text{urine}}}{m_{\text{lake}}+m_{\text{urine}}}\times (T_{\text{urine}}-T_{\text{lake}})mlake+murinemurine×(Turine−Tlake). Fraction = 4.0×1091.21×1016+4.0×109≈3.3059×10−7 \dfrac{4.0\times10^{9}}{1.21\times10^{16} + 4.0\times10^{9}} \approx 3.3059\times10^{-7}1.21×1016+4.0×1094.0×109≈3.3059×10−7. ΔT = fraction × 33.0 K ≈ 3.3059×10−7×33.0≈1.09×10−5 K3.3059\times10^{-7}\times33.0 \approx 1.09\times10^{-5}\ \text{K}3.3059×10−7×33.0≈1.09×10−5 K.

So the lake would warm by about 1.09×10⁻⁵ °C, i.e. ≈ 11 microkelvin.