tags: #publish links: [[My Writing]] created: 2021-08-10 Tue --- # How much energy would it take to boil the ocean? A common piece of advice when solving a problem is to avoid overdoing things and *trying to boil the ocean*. (Presumably, contrasted with just warming it up a bit, which is a more desirable outcome[^climate].) If you are a person working in the tech industry: in order to develop a good sense of the amount of time worth investing in different tasks, **it is useful to know how much energy is needed to boil the ocean, to help calibrate your sense of proportion**. ## Mass of ocean water Volume of oceans: 1.335 × 10^9 cubic kilometres × 10^9 m3 per km3 = Volume of oceans: 1.335 x 10^18 cubic metres × 10^3 litres per m3 = Volume of oceans: 1.335 × 10^21 litres × 10^3 g per litre = **Mass of ocean:** 1.335 × 10^24 grams of water. We're completely ignoring the dissolved salt content - we'll just heat the water. ## How many degrees do we need to heat it? Deep ocean water is mostly between -2 and 5°C. A thin layer of surface water is warmer depending on location, so let's assume it's 5°C on average, which is likely an overestimate. It's better to underestimate the heat input required: we don't want to overheat the ocean; our goal is to boil it, not evaporate it! Let's also assume that ocean water boils at 100°C. It actually boils higher, because of the salt. The water down in the depths of the ocean trenches boils *much* higher, because of the colossal pressure - in fact there are ocean floor thermal vents with liquid water at hundreds of °C, and communities of life with exotic biochemistry in the hot darkness. But we don't care about that! We just need to get the surface water boiling, and it's at atmospheric pressure. At least we don't have to account for the effect of altitude on boiling point, since we're at sea level! **Temperature increase required to get the water to boiling point:** 95°C ## How much energy is that? Mass of ocean: 1.335 × 10^24 grams × 4.2 J/g/°C × temperature increase of 95°C = **Energy required to boil the ocean**: 5.3 × 10^26 J. What's that in intuitive terms? ### Relative to world annual electricity production Annual production is about 27,000 TWh (2020). 27,000 TWh / year × 10^12 W x 3,600 seconds = World electricity production: 9.36 × 10^19 J/year 5.3 × 10^26 J to boil the ocean / 9.36 × 10^19 J/year = **about 5.5 million years worth of world electricity production** at 2020 rate. ### Relative to your kettle Assuming you have one of those chunky 2.4kW kettles: 5.3 × 10^26 J to boil the ocean / 2,400 J/s from your kettle = **7 × 10^15 years of leaving the kettle on in the sea**. ### Relative to solar output The sun produces around 3.8 × 10^26 W, mostly as radiation. 5.3 × 10^26 J to boil the ocean / 3.8 × 10^26 J/s = **about 1.4 seconds of total solar output to boil the ocean**. ## Conclusion So, next time someone uses this analogy, point out that *if a problem seems hard, you may be trying to solve the wrong problem*. **Instead of trying to figure out how to boil the ocean, *simply work on capturing and redirecting total solar output instead*.** Once that is solved - which is useful in other ways - then the original problem takes only a few moments to solve. [^climate]: Is it in fact a desirable outcome? Actually, no. IPCC tells us this leads to extreme weather events and sea level rise, so is generally quite bad for almost everybody.