# estimate

## 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:

• The volume of Lake Washington is 2.89 km31
• The average lake temperature is 9.71°C2
• It takes 4.19 Joules to raise the temperature of 1g of water by 1°C: $\frac{4.19 \mathrm{J}}{ 1 \mathrm{g^{\circ}C \: _{H_{2}O}}}$3
• The density of water is $\frac{1 \mathrm{g}}{1\mathrm{cm^{3}}}$

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}$4

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 Washington5.

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}}$6. “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.”7

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}}$8

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 $\frac{.625\times10^{18}\mathrm{J}}{1.74 \times 10^{17} \mathrm{\frac{J}{s}}} = 43.82 \: \mathrm{seconds}$ 9.

0
1. http://wldb.ilec.or.jp/Lake.asp?LakeID=NAM-09&RoutePrm=0:;14:load;14:load;

2. Average of all temperature data for 2011 for the Lake Washington buoy: http://green.kingcounty.gov/lake-buoy/Data.aspx

3. ROM estimate

4. this is why I’m not a chemist

5. still a ROM estimate