Help with Energy unit Joules EASY

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Homework Help Overview

The discussion revolves around calculating the energy released by the sun using the formula E = mc², where the mass of matter converted to energy is given as 3.6 x 10⁹ kg per second. Participants are exploring the relationship between energy and units, particularly Joules.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to calculate energy using the mass-energy equivalence formula and expresses confusion about unit conversion to Joules. Participants provide hints about the definition of a Joule and its relation to Newtons and kilograms.

Discussion Status

Participants are actively discussing unit conversions and confirming the calculations related to energy in Joules. Some guidance has been offered regarding the definitions of units, and there is a sense of progress as the original poster seeks validation of their calculations.

Contextual Notes

The original poster references a textbook definition of c² in terms of Joules per kilogram, indicating a potential point of confusion regarding unit consistency. There is also mention of natural units in sub-atomic physics, suggesting a broader context for the discussion.

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[SOLVED] Help with Energy unit ... Joules... EASY!

Homework Statement



The sun burns up 3.6 x 10^{9} kg of matter each second.
Using the formula E = mc^{2}, determine how much energy the
sun releases each second. [Speed of light: 3 x 10^{8} m/s]


Homework Equations



E = mc^{2}


The Attempt at a Solution



E = mc^{2}
. .= (3.6 x 10^{9} kg)(3 x 10^{8} m/s)^{2}
. .= (3.6 x 10^{9} kg)(9 x 10^{16} m^{2}/s^{2})
. .= 32.4 x 10^{25} \frac{kg \cdot m^2}{s^2}



Now I'm stumped with what to do with the units... I know it should be Joules per second. But how do I show in my solution that Joules will come out? Joules is N\cdotm!

Do I have to change the kg into N? ... My textbook says that the magnitude of c^{2} is 9 x 10^{16} Joules per kilogram.

Please check my calculations also!
Thank You! :smile:
 
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First hint: How is "Joule" defined?
Second hint: If E = mc^{2} is valid, and you have SI units to the right, then you must obtain SI units on the left hand side as well.

The second hint would not be true if you was working with the natural units in sub-atomic phyiscs, where both c and h-bar = 1 i.e E = m, then one must do a more careful unit analysis.
 
A Joule is N m and a Newton, N, is kg m/s2. So a Joule is kgm2/s2. That's exactly what you have!
 
Hmmm...

well Joule is the unit of work. N*m
Newton x meter
 
HallsofIvy said:
A Joule is N m and a Newton, N, is kg m/s2. So a Joule is kgm2/s2. That's exactly what you have!


So that's understood? I'll then just add...

E = mc^{2}
. .= (3.6 x 10^{9} kg)(3 x 10^{8} m/s)^{2}
. .= (3.6 x 10^{9} kg)(9 x 10^{16} m^{2}/s^{2})
. .= 32.4 x 10^{25} \frac{kg \cdot m^2}{s^2}
. .= 32.4 x 10^{25} Joules

Simple as that?
Yipee! :biggrin:

Are my calculations correct? exponents...
 
Yes, since F [N] = m*a [kg * m/s^2 ]
 

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