Answer
(a) $4.33\times 10^9Kg/s$
(b) $0.0307\%$
Work Step by Step
(a) We can find the required rate of conversion as
$\frac{\Delta m}{\Delta t}=\frac{P_{sum}}{c^2}$
We plug in the known values to obtain:
$\frac{\Delta m}{\Delta t}=\frac{3.90\times 10^{26}}{3.00\times 10^8}$
$\frac{\Delta m}{\Delta t}=4.33\times 10^9Kg/s$
(b) We know that
$\Delta M=\frac{\Delta m}{\Delta t}T$
We plug in the known values to obtain:
$\Delta M=(4.33\times 10^9Kg/s)(4.5\times 10^9y)(\frac{3.16\times 10^7s}{1y})$
$\Delta M=6.16\times 10^{26}Kg$
Now $\frac{\Delta M}{M+\Delta M}=\frac{6.16\times 10^{26}}{2.00\times 10^{30}+6.16\times 10^{26}}\times 100=0.0307\%$
