Answer:
[tex]5.2416\times 10^8 kg[/tex]kilograms of sulfur will be emitted from coal plants in two weeks
Explanation:
Power produced by power plants in the United States :P
P=[tex]2.6\times 10^{11} Watts [/tex]
1 Joule = Watts × Seconds
Watts = Joule/Second
P=[tex]2.6\times 10^{11} J/s[/tex]
This means that [tex]2.6\times 10^{11} [/tex] Joules of enrgy is produced in one second.
So, energy produced in 2 week:
1 week = 7 days
1 day = 24 hours
1 hour = 3600 seconds
2 week = 2 × 7 × 24 × 3600 s=1,209,600 s
Energy produced in 1,209,600 s: E
[tex]E=2.6\times 10^{11} J/s\times 1,209,600 s[/tex]
Coal has an energy content =[tex]2.4\times 10^7 J/Kg[/tex]
Mass of coal used while producing E amount of energy: m
[tex]m\times 2.4\times 10^7 J/Kg=E[/tex]
[tex]m=\frac{E}{2.4\times 10^7 J/Kg}[/tex]
[tex]m=\frac{2.6\times 10^{11} J/s\times 1,209,600 s}{2.4\times 10^7 J/Kg}[/tex]
[tex]m = 1.3104\times 10^{10} kg[/tex]
Percentage of sulfur in coal = 4% by mass
Mass of sulfur produced from [tex]1.3104\times 10^{10} kg[/tex] in 2 weeks:
[tex]\frac{4}{100}\times 1.3104\times 10^{10} kg=5.2416\times 10^8 kg[/tex]
