Chapter 29 - Relativity - Problems and Conceptual Exercises - Page 1043: 63

$0.35Kg$

We know that $E=Pt$ $\implies E=(1.0\times 10^9)(3.16\times 10^7)=3.16\times 10^{16}J$ We also know that $E=\Delta mc^2$ This simplifies to: $\Delta m=\frac{E}{c^2}$ We plug in the known values to obtain: $\Delta m=\frac{3.16\times 10^{16}}{(3.0\times 10^8)^2}$ $\Delta m=0.35Kg$