One day, as the soothing warmth of earl grey tea trickled down my throat, I wondered just how much power the electric kettle consumed in boiling a half-liter bottle's worth of water, so I crunched some numbers...
Recall that Power (P) is the time rate of change of energy--which in this case is the work done by the electric boiler on the water.
$$
P = \frac{dW}{dt}
$$
The work done by the boiler is equivalent to the change in internal energy (U) of the water as the water heats up to 100°C. This change in internal energy is directly proportional to the change in temperature at constant pressure.
$$
dW = dU = mC_{p}dT
$$
This allows us to relate power to the change in temperature of the water.
$$
P = mC_{p}\frac{dT}{dt}
$$
Since we're not working with infinitesimal changes, we can rewrite this relationship in terms of simple changes of temperature and time.
$$
P = mC_{p}\frac{\Delta T}{\Delta t}
$$
Now let's plug in some numbers!
$$\Delta T = 100^{\circ}C - 23^{\circ}C = 77^{\circ}C \equiv 77^{\circ}K \\
\Delta t = 3 min \times \frac{60 s}{1 min} = 180s \\
\rho_{H_{2}O} \approx 1 \frac{kg}{L} \therefore m = V \times \rho_{H_{2}O} = 0.5L \times \frac{kg}{L}= 0.5 kg\\
C_{p} = 4.1813 \frac{kJ}{kg^{\circ}K}$$
$$
P = 0.5 kg \times 4.1813 \frac{kJ}{kg^{\circ}K} \times \frac{77^{\circ}K}{180s} \approx 0.894 \frac{kJ}{s} \\
0.894 \frac{kJ}{s} \equiv 0.894 \ kW \equiv 894 W
$$
So my roommate's electric kettle consumes 894 Watts of power. However, that doesn't mean much, so let's dig deeper...
According to the Energy Information Administration, America generates 39% of its electricity from coal. Coal has an energy density of 32,000 kilo-Joules per kilogram.
Well we already implicitly calculated the kettle's energy consumption to be about 160 kJ (recall P = W/t, so W = Pt). Let's use that to calculate my coal consumption.
$$
m_{coal} = W \div \rho_{energy_{coal}} = 160 kJ \div 32,000 \frac{kJ}{kg} = 0.005 kg \equiv 5g
$$
Hmm, 5 grams doesn't look like much, but let's scale that up by a conservative estimate of 10% of the US population, or 31.6 million people.
$$
m_{coal_{US}} = 0.005 kg \times 31.6 \times 10^{6} = 158,000 kg \\
m_{coal_{US}} = 158,000 kg \times \frac{1t}{1000 kg} = 158 t
$$
158 tonnes! That's the mass equivalent of 22 elephants! That means 10 million Americans boiling half a kilo of water for a cup of tea use up 158 tonnes of coal--every day!
What if we used another power source? Well when I looked up the mass-energy density of coal on Wikipedia, I noticed the immense mass-energy density of Uranium as used in a breeder reactor.
$$
\rho_{energy_{Uranium}} = 80,620,000 \frac{MJ}{kg}
$$
Wow! That's more than 3 million times as energy dense as coal! Let's recalculate the fuel-mass required for boiling water.
$$
m_{fuel} = m_{coal_{US}} \times \frac{ \rho_{ energy_{coal} } } {\rho_{energy_{Uranium}}} \\
m_{fuel} = 158,000 kg \times \frac{32}{80,620,000} \approx 0.063 kg \equiv 63 g
$$
63 grams. Compare the fiscal and environmental costs of mining, refining, and storing a mere 63 grams of Uranium to the fiscal and environmental costs of mining, refining, and burning of 158 thousand kilograms of coal next time you think about atomic power. Yes, nuclear power plants are dangerous if not carefully supervised, but the near-unlimited power that we can derive from nuclear plants far outweighs their cost.
And speaking of unlimited power, let's conclude with the best thing out of the Star Wars prequels...
