The sun radiates energy at the rate of approximately 3.80 x 1026 W. Assume that this energy is produced by a reaction whose net result is the fusion of four protons to form a single 4He nucleus and the release of 25.0 MeV of energy that is radiated into space. Calculate the sun's loss of mass per day.
You only need the first fact, and the equivalence between mass and energy introduced by Einstein, E = mc^2. Since you're interested in mass loss per unit time, you can take the derivative of both sides: \[\frac{dE}{dt} = c^2 \frac{dm}{dt}\] Or, since dE/dt is the power P: \[\frac{dm}{dt} = \frac{P}{c^2}\] I get 4.23 x 10^9 kg/s, or about 1 trillionth of 1% of the mass of the Sun per year.
great! I was just confused as to why I was given two energy values
Your answer is correct though, so thank you very much! I really appreciate it
It's routine in more sophisticated questions to give you more data than you need. It's called "adding distractors." The theory is that it better tests whether you understand the principle, because you have to choose which data to use.
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