# The Implausibility of the Greenhouse Effect On Ocean Temperature

According to an Environmental Protection Agency (EPA) whitepaper from 2016, total ocean heat content has grown since 1970 by between 100 and 200 zettajoules (10^21). The National Aeronautics and Space Administration (NASA) currently lists it at over 320 zettajoules into 2020.

While that’s a massive amount of heat, the oceans are also massive in size so it hasn’t affected ocean temperatures substantially. Water also has a high heat capacity; it takes more energy to raise the temperature of water than it does air.

According to NASA, oceans have absorbed 90 percent of the warming from the greenhouse effect—gasses like carbon dioxide, methane, and nitrogen oxide in the atmosphere absorbing heat as a greenhouse would, rather than allowing that heat to escape.

But if that additional heat in the ocean is coming from atmospheric gases, that would imply an incredibly massive and unlikely transfer of energy into the oceans through the atmosphere that would likely affect air temperatures along the way.

Because of the differences in heat capacity and density, air is more sensitive to heat than water. The heat capacity of air is 700 Joules per kg per degree Kelvin, while for water it is around six times that (4,184 J/kgK). Water has a density of 1,000 kg/m³ while air is just 1.2 kg/m³.

If that heat passed through the atmosphere, it would lead to a temperature increase of over 2.25 °Celsius (4.05 °Fahrenheit)—not the .89 °Celsius (1.6 °Fahrenheit) of warming since 1970 shown in NASA data.

And that is assuming the smallest measure of increased heat from Japan Meteorological Agency’s Meteorological Research Institute (MRI/JMA) (100 zettajoules) seen in the EPA’s paper.

The higher value from NASA (320 zettajoules) would lead to a change in atmospheric temperature over three times higher (7.31 °Celsius/13.16 °Fahrenheit).

This is a broad estimate that doesn’t account for many complexities of climate activities but serves as a scale of what such a large influx of heat in the atmosphere would do.

In general, there’s no reason why the atmosphere would pass on 90 percent of the added heat into the oceans—as described by NASA—rather than a much smaller amount based on surface area.

## Calculation of Air Temperature Change

The total thermal energy stored in a substance is defined by the equation:

Energy = Volume x Density x Temperature x Heat Capacity

And the equation for the change in energy would be similar:

Change in Energy = Volume x Density x Change in Temperature x Heat Capacity

Solving for the Change in Temperature, we get:

Change in Temperature = Change in Energy / (Volume x Density x Heat Capacity)

For air, the inputs are:

##### Change in Energy Since 1970 (EPA data, NASA data)

100-300 zettajoules

##### Density of Air (source)

1.293 kg/m³ = 1.293 * 10^9 kg/km³

##### Heat Capacity of Air (source)

700 Joules per kg per degree Kelvin

##### Volume

Based on numbers from NASA, Earth’s radius is 4,000 miles, and the atmosphere is 60 miles (~100 km) above that—otherwise known as the Kármán line.

The volume of the atmosphere is then equal to the volume of Earth including the atmosphere minus the volume of Earth. With the equation for volume (4/3 x π x r³), we can then calculate the volume of the atmosphere:

Volume of Atmosphere = Volume of Earth and Atmosphere - Volume of Earth

Volume of Atmosphere =(4/3 x π) x ((Earth Radius + Atmosphere Height)³) - (Earth Radius³))

The equation inputs are:

Earth Radius = 4,000 miles = 6,437.376 kilometers

Atmosphere Height = 100 kilometers

Volume of Atmosphere = (4/3 x π x (6,437.376 km + 100 km)³) - (π x (6,437.376 km)³)

Volume of Atmosphere = 52,887,934,086 km³ = ~53 billion km³

##### Final Calculation

Entering those inputs into our equation for the change in temperature, and using the smallest possible value of total heat transfer, we get:

Change in Temperature = Change in Energy / (Volume x Density x Heat Capacity)

Change in Temperature (Kelvin) = 100 zettajoules / (53 billion km³ x 1.2 * 10^9 kg/km³ x 700 Joules per kg per degree Kelvin)

Change (Δ) in Temperature = 2.25 Δ°Kelvin = 2.25 Δ°Celsius = 4.05 Δ°Fahrenheit