## College Physics (4th Edition)

$E = 2.0\times 10^{47}~J$
We can find the energy released by the explosion: $E = mc^2$ $E = (0.80)(1.4)(m_{sun})~c^2$ $E = (0.80)(1.4)(1.989\times 10^{30}~kg)(3.0\times 10^8~m/s)^2$ $E = 2.0\times 10^{47}~J$