Question 28: Atoms & Nuclei

Question

To get 300 MW electric power for half an hour, how much mass is to be completely converted into energy?

Accepted Answer

Let
P = 300 MW = 300ÃÂ10^6 W
t = ÃÂ½ h = 1800 s
Energy required:
$$E = P\,t = (300	imes10^6)\,(1800) = 5.4	imes10^{11}	ext{ J}$$
Mass fromÃ¢ÂÂ$$E = mc^2\;\Longrightarrow\;m=\frac{E}{c^2}$$
$$m = \frac{5.4	imes10^{11}}{(3	imes10^8)^2}         = \frac{5.4	imes10^{11}}{9	imes10^{16}}         = 6	imes10^{-6}	ext{ kg}$$
So the required mass is 6ÃÂ10Ã¢ÂÂ»Ã¢ÂÂ¶ kg.
Answer: Option D.

Answer

$6 	imes 10^{-2} \mathrm{~kg}$

Answer

$3 	imes 10^{-6} \mathrm{~kg}$

Answer

$6 	imes 10^{-3} \mathrm{~kg}$

Answer

$6 	imes 10^{-6} \mathrm{~kg}$

Atoms & Nuclei

To get 300 MW electric power for half an hour, how much mass is to be completely converted into energy?

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