Posted on by Andrew Ferguson

Vaporizing Lake Washington

Sometimes I wonder about interesting things, such as how much energy would it take to boil all the water of Lake Washington:

Putting all that together, we get:

2.89 \mathrm{km^{3}} \times \left ( \frac{1000\mathrm{m}}{1\mathrm{km}} \right )^{3} \times \left( \frac{100\mathrm{cm}}{1\mathrm{m}} \right )^{3} \times \frac{1\mathrm{g_{_{H_{2}0}}}}{1\mathrm{cm^{3}_{H_{2}0}}} \times \frac{4.19 \mathrm{J\:} }{\mathrm{g^{\circ} C _{H_{2}O}}} \times \left ( 100^{\circ} \mathrm{C} - 9.71^{\circ} \mathrm{C}\right )= 1.093\times10^{18}\mathrm{J}

For comparison, the energy that hits Earth from the Sun in one second: 1.74 \times 10^{17} \mathrm{J} ((According to Wolfram Alpha))

Basically, if we could focus all the energy from the sun that hits the earth, it would take…\frac{1.093\times10^{18}\mathrm{J}}{1.74 \times 10^{17} \mathrm{\frac{J}{s}}} = 6.281 \: \mathrm{seconds} …to vaporize Lake Washington ((ROM estimate)).

This is a vast oversimplification of the forces and energies involved, but I think it’s still a pretty good estimate.

Update: Apparently I missed one critical element, enthalpy/heat of vaporization \Delta{}H_{\mathrm{vap}} ((this is why I’m not a chemist)). “This energy breaks down the intermolecular attractive forces, and also must provide the energy necessary to expand the gas (the PΔV work). For an ideal gas , there is no longer any potential energy associated with intermolecular forces. So the internal energy is entirely in the molecular kinetic energy.” ((http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/phase2.html#c3))

What we have above is the energy required to bring it up to 100°C, but not to vaporize it. To actually vaporize water that’s already at 100°C, we need to add an additional \Delta{}H_{\mathrm{vap}} = 2260\mathrm{\frac{J}{g}} ((http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/phase.html))

Running this number back through our calculations, we now get:
2.89 \mathrm{km^{3}} \times \left ( \frac{1000\mathrm{m}}{1\mathrm{km}} \right )^{3} \times \left( \frac{100\mathrm{cm}}{1\mathrm{m}} \right )^{3} \times \left (2260\mathrm{\frac{J}{g}} + \frac{1\mathrm{g_{_{H_{2}0}}}}{1\mathrm{cm^{3}_{H_{2}0}}} \times \frac{4.19 \mathrm{J\:} }{\mathrm{g^{\circ} C _{H_{2}O}}} \times \left ( 100^{\circ} \mathrm{C} - 9.71^{\circ} \mathrm{C}\right ) \right ) = 7.625\times10^{18}\mathrm{J}

This is still within one order of magnitude from my original answer and really only takes slightly longer for the sun to actually vaporize Lake Washington \frac{.625\times10^{18}\mathrm{J}}{1.74 \times 10^{17} \mathrm{\frac{J}{s}}} = 43.82 \: \mathrm{seconds} ((still a ROM estimate)).

10 Replies to “Vaporizing Lake Washington”

Leave a Reply

Your email address will not be published. Required fields are marked *