I’ll be traveling a bit for the next two weeks, so my posts may be a little sporadic. When I get back, I’ll try to address all the questions people have left in the comments and I’ll post a few of the Fermi questions readers have sent me. In the mean time, I’m going to try to answer one of Thomas’s questions from a while back:
If we took a cubic meter of the sun and dropped it into the ocean, how much water would boil away?
As a disclaimer, I know very little about astrophysics, plasma physics, and nuclear physics. That said, I think I can still come up with a decent estimate even with my limited knowledge. I’m assuming Thomas means a cubic meter of the Sun’s core, which is both denser and hotter than its surface. According to Wikipedia, the Sun’s core has a temperature of 1.57×107 K and a density of 1.5×105 kg/m3. The core is too hot for atoms to be stable, so it’s basically a bunch of free protons, neutrons, and electrons swishing about. Physicists call this phase of matter “plasma.” When we dump a cubic meter of this plasma into the ocean, it will cool as it gives off heat to the surrounding water. As it cools, the protons will bind with electrons to form atoms—I’ll assume they all form hydrogen atoms—and these hydrogen atoms will then form hydrogen molecules. Each of these phase changes releases some heat1 into the water. To compute how much energy is given to the water, we need to know both the energy gained by cooling and the energy gained by phase changes.
From the density, we know that the total mass of one cubic meter of the Sun’s core is about 1.5×105 kg. Since the atomic weight of a hydrogen atom is 1.008 g/mol, we can compute the total number of hydrogen atoms we’ll get out of the chunk o’ Sun,
number of atoms = (mass) / (atomic weight)
= (1.5×105 kg) / (1.008 g/mol)
= 1.5×105 mol.
This is equivalent to 9.0×1031 hydrogen atoms. The ground state of a hydrogen atom is –1300 kJ/mol (~-13.6 eV). From this we can compute the amount of energy released in the formation of hydrogen atoms,
energy released = (energy per atom) · (# of atoms)
= (1300 kJ/mol) · (1.5×105 mol)
= 2.0×108 kJ.
It takes two hydrogen atoms to form a hydrogen molecule. This means the 9.0×1031 hydrogen atoms in our glob will form 4.5×1031 molecules or 7.5×104 moles of molecules. Each H2 molecule has a binding energy of 436 kJ/mol, so the total heat released by the formation of molecules is
energy released = (energy per molecule) · (# of molecules)
= (436 kJ/mol) · (7.5×104 mol hydrogen molecules)
= 3.3×107 kJ.
There’s also energy released in the cooling process. I’ll assume both the plasma and molecular hydrogen are well approximated by an ideal gas, which has a specific heat of 20.79 J/K·mol at constant pressure2. From this, the temperature change, and the number of moles of hydrogen atoms, we can estimate the heat released by cooling,
energy released = (specific heat) · (# of atoms) · (temperature change)
= (20.79 J/K·mol) · (1.5×105 mol) · (1.57×107 K)
= 4.9×1010 kJ.
As you can see, the heat released by cooling is much larger than either of the phase changes indicating that this is the only significant factor. This means about 4.9×1010 kJ of energy is used to boil ocean water.
To determine how much water boils away, we need to know water’s specific heat (~4.186 kJ/kg·K) and its latent heat of vaporization (~2260 kJ/kg). If the ocean starts at room temperature (~298 K = 25 °C), it will have to increase by about 75 K in order to boil. Using this data, we can estimate the total mass of water that would be boiled away3.
mass = (energy added) / [(latent heat) + (specific heat) · (temperature change)]
= (4.9×1010 kJ) / [(2260 kJ/kg) + (4.186 kJ/kg·K) · (75 K)]
= 1.9×107 kg.
It would boil away about 20 million kilograms of water. That’s about a cube of water 27 m on a side.
 Physicists call this “latent heat.”
 The specific heat of an ideal gas at constant pressure is equivalent to 5R/2, where R is Boltsmann’s constant. For simplicity, I’m assuming all the heat gets transferred directly to the water that gets boiled off (i.e. the rest of the ocean stays at room temperature.)