Q83PE - College Physics (Urone)

Question

Construct Your Own Problem

Consider the decay of radioactive substances in the Earth's interior. The energy emitted is converted to thermal energy that reaches the earth's surface and is radiated away into cold dark space. Construct a problem in which you estimate the activity in a cubic meter of earth rock? And then calculate the power generated. Calculate how much power must cross each square meter of the Earth's surface if the power is dissipated at the same rate as it is generated. Among the things to consider are the activity per cubic meter, the energy per decay, and the size of the Earth.

Step-by-Step Solution

Answer

a) The activity of the earth rock in cubic meter is \(R = \frac{E}{{{E_\alpha }}}\).

b) Calculated power generate is \(P\) is \(\left( {1.6 \times {{10}^{ - 13}}} \right)q\;W\).

c) The power crosses per unit cubic meter is \(\left( {1.48 \times {{10}^{ - 33}}} \right)q\frac{{\rm{W}}}{{{{\rm{m}}^3}}}\).

Definition of thermal energy

The energy contained within a system that is accountable for its temperature is referred to as thermal energy. The transfer of thermal energy is referred to as heat.

Activity of the earth rock in cubic meter

Consider the given problem.

The energy, the heat radiated and the volume of earth is related by the formula given by:

\(\begin{align}{\underline{\phantom{xx}}}E & = \frac{Q}{{{V_e}}}\\E & = \frac{{q\,{\rm{MeV}}}}{{\frac{4}{3}\pi {{\rm{R}}^3}}}\end{align}\)

Where \(Q\) is the heat radiated, \(E\) is energy and the \({V_e}\) is the volume of earth

Now substitute the value of \(R & = 6380\;{\rm{km}}\) as the radius of earth is \(6380\;{\rm{km}}\) in the above equation

\(\begin{align}{\underline{\phantom{xx}}}E & = \frac{{q\,{\rm{MeV}}/{\rm{s}}}}{{\frac{4}{3}\pi {{(6380\,{\rm{Km}})}^3}}}\\ & = \frac{{q\,{\rm{MeV}}/{\rm{s}}}}{{\frac{4}{3} \times 3.14{{\left( {6380 \times {{10}^3}\;\,{\rm{m}}} \right)}^3}}}\\ & = \frac{{q\,{\rm{MeV}}/{\rm{s}}}}{{1.09 \times {{10}^{21}}\;\,{{\rm{m}}^3}}}\end{align}\)

Thus, the activity of earth's rock is given by:

\(R & = \frac{{\rm{E}}}{{{{\rm{E}}_\alpha }}}\)

 

Therefore, the value of \(R\) is \(\frac{{\rm{E}}}{{{{\rm{E}}_\alpha }}}\).

Calculate the power generated

b)

Find the power generated.

The power generating is the energy per unit time \(P = E\).

so after plug in the value we get,

\(\begin{array}{c}P = q\,{\rm{MeV}}/{\rm{s}}\\ = \left( {1.6 \times {{10}^{ - 13}}} \right)\,q\;{\rm{J}}/{\rm{s}}\\ = \left( {1.6 \times {{10}^{ - 13}}} \right)\,q\;{\rm{W}}\\P = \left( {1.6 \times {{10}^{ - 13}}} \right)\,q\;{\rm{W}}\end{array}\)

Therefore, the value of \(P\) is\(\left( {1.6 \times {{10}^{ - 13}}} \right)q\;{\rm{W}}\).

The power crosses per unit cubic meter

c)

Let us solve the given problem. 

The power crosses per unit cubic meter is \(p = \frac{{\rm{P}}}{{{{\rm{V}}_e}}}\).

Plug in the value we get ,

\(\begin{array}{c}p = \frac{{\left( {1.6 \times {{10}^{ - 13}}} \right)q\;{\rm{W}}}}{{1.09 \times {{10}^{21}}\;{{\rm{m}}^3}}}\\ = \left( {1.48 \times {{10}^{ - 33}}} \right)q\,\frac{{\rm{W}}}{{{{\rm{m}}^3}}}p\\ = \left( {1.48 \times {{10}^{ - 33}}} \right)q\,\frac{{\rm{W}}}{{{{\rm{m}}^3}}}\end{array}\)

Therefore, the value of \(t\) is \(\left( {1.48 \times {{10}^{ - 33}}} \right)q\frac{{\rm{W}}}{{{{\rm{m}}^3}}}\).