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The total mass of the Sun is about 2×10^30 kg, of which about 76 % was hydrogen when the Sun formed. However, only about 12 % of this hydrogen ever becomes available for fusion in the core. The rest remains in layers of the Sun where the temperature is too low for fusion.
Part A
Use the given data to calculate the total mass of hydrogen available for fusion over the lifetime of the Sun.
Express your answer using two significant figures.
Part B
The Sun fuses about 600 billion kilograms of hydrogen each second. Based on your result from part A, calculate how long the Sun’s initial supply of hydrogen can last. Give your answer in both seconds and years.
Express your answer using two significant figures.
Part D
Given that our solar system is now about 4.6 billion years old, when will we need to worry about the Sun running out of hydrogen for fusion?
Express your answer using two significant figures.

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Answer:

A. 1.8 ×[tex]10^{30}[/tex] Kg

B i. 3.0 × [tex]10^{17}[/tex] seconds

  ii. 9.6 × [tex]10^{9}[/tex] years

C. After 9.2 × [tex]10^{9}[/tex] (9.2 billion) years

Step-by-step explanation:

Given that the mass of the Sun = 2× [tex]10^{30}[/tex] Kg.

Mass of hydrogen when Sun was formed = 76% of 2× [tex]10^{30}[/tex] Kg

                            = [tex]\frac{76}{100}[/tex]  ×2× [tex]10^{30}[/tex] Kg

                           = 1.52 × [tex]10^{30}[/tex] Kg

Mass of hydrogen available for fusion = 12% of 1.52 × [tex]10^{30}[/tex] Kg

                           = [tex]\frac{12}{100}[/tex] × 1.52 × [tex]10^{30}[/tex] Kg

                           = 1.824 ×[tex]10^{30}[/tex] Kg

A. Total mass of hydrogen available for fusion over the lifetime of the sun is 1.8 ×[tex]10^{30}[/tex] Kg.

B. Given that the Sun fuses 6 × [tex]10^{11}[/tex] Kg of hydrogen each second.

i. The Sun's initial hydrogen would last;

                                     [tex]\frac{1.8*10^{30} }{6*10^{11} }[/tex]

                                 = 3.04 × [tex]10^{17}[/tex] seconds

The Sun's hydrogen would last 3.0 × [tex]10^{17}[/tex] seconds

ii. Since there are 31536000 seconds in a year, then;

The Sun's initial hydrogen would last;

                                     [tex]\frac{3.04*10^{17} }{31536000}[/tex]

                                 = 9.640 × [tex]10^{9}[/tex] years

The Sun's hydrogen would last 9.6 × [tex]10^{9}[/tex] years.

C. Given that our solar system is now about 4.6 × [tex]10^{9}[/tex] years, then;

                               [tex]\frac{9.6*10^{9} }{4.6*10^{9} }[/tex]

                             = 2.09

So that;   2 × 4.6 × [tex]10^{9}[/tex] = 9.2 × [tex]10^{9}[/tex] years

Therefore, we need to worry about the Sun running out of hydrogen for fusion after 9.2 × [tex]10^{9}[/tex] years.

Part(A): The total mass of hydrogen available 9.6 billion years.

Part(B): The total time is 5.10 billion years.

Part(D): The hydrogen will last [tex]5.04\times 10^9 \ years[/tex]

Mass of the sun:

Our sun is the largest object in our solar system. The mass of the sun is approximately [tex]1.988\times 1030[/tex] kilograms

Part(A):

Given that,

The total mass of the Sun =[tex]2\times10^{30} kg[/tex]

Mass of hydrogen in Sun =  [tex]2\times10^{30} \times0.76\ kg[/tex]

The mass of hydrogen ever available for fusion is,

[tex]2\times10^{30} \times 0.76 \times 0.12 kg = 1.824\times 10^{29}[/tex]

Mass of hydrogen fuses each second = 600 billion kg/second.

Time hydrogen will last in seconds=[tex]1.824\times 10^{29}seconds.[/tex]

[tex]= 0.00304\times 10^{29 }= 3.04\times 10^{17}.[/tex]

Time hydrogen will last in seconds =[tex]1 year = 31,536,000 seconds.[/tex]

[tex]31,536,000 x = 3.04\times 10^{17} = 9.6 \ billion \ years.[/tex]

(B) Present age of sun = [tex]9.6-4.5 \ billion \ years[/tex]

The time when we need to worry about Sun running out of hydrogen for fusion = 5.10 billion years.

(D) The solar system is 4.6 billion years old that is [tex]4.6\times 10^9\ years[/tex]

And in part (B) we have calculated that hydrogen will last [tex]9.64\times 10^9[/tex] then,

[tex]9.64\times 10^9-4.6\times 10^9=5.04\times 10^9[/tex]

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