Archived Estimate nuclear energy generation rate (formula given) per unit MASS MASS?

Click For Summary
The discussion revolves around estimating the nuclear energy generation rate for the proton-proton (pp) cycle in the Sun using a provided formula. The user correctly substituted values into the formula and calculated an energy generation rate of 3.186 x 10^-4 W kg^-1. However, confusion arises regarding the phrase "per unit mass mass," which seems to be a typographical error. Clarification indicates that W/kg is appropriate for the problem statement, suggesting that the user’s calculation is indeed correct. The focus remains on understanding the application of the formula in the context of stellar fusion processes.
daleklama
Messages
33
Reaction score
0

Homework Statement



Consider energy generation in the Sun. Fusion through the pp cycle takes place within its core which has

r = 0.2 Rs
T = 1.3 x 10 ^ 7 K
rho (density) = 1.1 x 10 ^ 5 kg m^-3
X (total mass fraction of hydrogen) = 0.34

The nuclear energy generation rate for the pp cycle (where T6 = T/10^6K and density has units of kg m ^ -3) is

Epp = (0.241) (rho) (X^2) (T6 ^ (-2/3) ) exp (-33.8/(T6 ^ 1/3)) W kg ^ -1

Estimate epp per unit mass mass.


Homework Equations



Epp = (0.241) (rho) (X^2) (T6 ^ (-2/3) ) exp (-33.8/(T6 ^ 1/3)) W kg ^ -1

The Attempt at a Solution



I subbed everything in:

Epp = (0.241) (1.1 x 10 ^ 5) (0.1156) ( 13 ^ -2/3) exp (-33.8/ (13 ^ 1/3))
(T6 is 13, because it says in the question that T6 = T/10^6K)

and I got 3.186 x 10 ^ -4 as my final answer, I assume the unit is W kg ^ -1 because it was given in the formula.

The bit I'm confused by is that in the question it specifies "estimate epp PER UNIT MASS MASS."

That bit I haven't a clue about... did I do it wrong?

Thanks for any help!
 
Physics news on Phys.org
W/kg is fine if the problem statement is "per unit mass" (or "per unit mass mass" which looks like a typo).
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
15
Views
2K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
10K
Replies
33
Views
8K