Answer to Question #160181 in Electrical Engineering for Keleko

Question

An all-electric home uses approximately 2000KWh of electric energy per month. How much uranium-235 would be required to provide this house with its energy needs for one year? Assume 100% conversion efficiency and 208 MeV released per fission.

Answer 1

Convert 2000 kW·h to J:

$$E=2000·1000·3600=7.2·10^9\text{ J}.$$

Find how many fissions (uranium atoms) can provide this amount of energy:

$$n=E/E_U=\frac{7.2·10^9}{208·1.6·10^{-13}}=2.16·10^{20}\text{ fissions}.$$

One uranium-235 weighs 235·1.66·10^-27 kg. So, the number of fissions we found above corresponds to the following mass of uranium:

$$M=nm=2.16·10^{20}·235·1.66·10^{-27}=8.43·10^{-5}\text{ kg}.$$