Openai/692b0130-6a64-800a-8103-b4de485f6aa0

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gpt-5-t-mini

User: if everyone on the planet peed into lake superior at the same time, how much would it increase the lake temp.

if everyone on the planet peed into lake superior at the same time, how much would it increase the lake temp.

Short answer

Almost zero — on the order of 10⁻⁵ °C (≈ 1.1×10⁻⁵ °C, about 11 microkelvin). Practically undetectable.

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

1. 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.
2. 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.
3. 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​+murine​mlake​Tlake​+murine​Turine​​. Temperature increase ≈ murinemlake+murine×(Turine−Tlake)\dfrac{m_{\text{urine}}}{m_{\text{lake}}+m_{\text{urine}}}\times (T_{\text{urine}}-T_{\text{lake}})mlake​+murine​murine​​×(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.

Sensitivity / sanity checks

• If average pee per person were 0.2 L instead of 0.5 L, the warming would be ≈ 4.4×10⁻⁶ °C.
• If it were 1.0 L each (very generous), warming ≈ 2.2×10⁻⁵ °C.
• Even in the extreme, this is utterly negligible compared with natural daily/seasonal temperature variations (which are degrees, not microdegrees).

Intuition / real-world note

Locally — right where millions of people stood — the water would be locally warmed slightly and chemically changed; but once mixed into the whole lake the effect disappears. In short: hilarious thought experiment, negligible climate or thermal effect.