Answers for Question Set for Sun’s Energy (First Draft 11/12/14 11am)
Solutions as of 11/15/14 7pm
1. Be able to calculate the energy released when a specific amount of hydrogen fuses to form Helium. Element masses are typically listed in terms of Atomic Mass Units and the conversion between AMU and Kg is 1 amu = 1.66053892173e-27 kilograms1. A periodic table is provided at the end of this sheet.
The basic conversion is 4H eventually fuse to form 1He. The total mass before is 4*1.0079amu (4.0316amu) and the mass after is 1*4.0026amu (4.0026amu).
Net loss is 0.029amu [(1.66054 x 10-27 kg) / 1 amu] = 4.8156 x 10-29kg
Using Einstein’s Principle of Equivalence (E = mc2) gives us an energy release of:
E = (4.8156 x 10-29kg)(3 x 108)2 = 4.33 x 10-12 Joules, which is a very small value but recall that you just did this for 4 atoms fusing to one!
2. Explain the conditions necessary for an H to He fusion reaction to occur and what happens.
If we start with the most basic assumptions, we will Hydrogen in abundance if we wish the reaction to yield any appreciable energy. To get atoms to fuse, they need to overcome the most basic of fundamental forces –the Strong Nuclear Force (sometimes called the Coulomb Barrier). The electrostatic repulsion between the protons of each element is staggering (my word choice) and the elements must be moving at incredibly high speeds to be able to collide. These speeds are directly related to the temperature of the elements, so high speeds = high temperatures. How high? How does about 10 million degrees grab you? This is essentially the exclusive domain of stars (nuclear furnaces fueled by fusion reactions. When fusion occurs, heavier elements are formed from the lighter constituent elements. In the process, some of the initial mass is converted into forms of pure energy (heat, light, etc.) via Einstein’s Equivalency Principle. In our case, 4 Hydrogen will fuse to form one Helium atom and a specific amount of energy.
3. Aside from very specific scientific experiments, why can’t fusion occur on Earth?
Nowhere on Earth, except for really small localized experiments that exist for fractions of a second (Fusion bombs, for example) is the temperature high enough (10 million K) to allow fusion to occur. Also, there isn’t a lot of free (unbound) hydrogen left on Earth since during the time of formation that gas was moving fast enough (KE) to escape the gravitational pull and head off for points unknown.
4. Explain the balance between competing forces that keep a relatively young a star relatively stable in energy output and size.
Due to the tremendous mass of stars in a relatively localized area, they essentially are being pulled inwards by their own gravitational attraction. The gravitational “pressure” is countered by the heat/energy “pressure” pushing outwards due to the nuclear fusion reactions converting H to He near Sun’s core. Yes Ghaida the mass of the sun is decreasing as the output of energy stays constant, so the outward pressure will eventually cause the sun’s outer “edge” to expand in size until at a point 5 billion years in the future, it will be where we are (where we were…).
5. If you look over the PV lab, you should be able to handle all the associated calculations including but not limited to: power emitted per square meter of a source at a predefined distance from that source, PV cell output based on current and voltage readings, the efficiency of a solar panel as we did in the PV lab, the effect of angle on PV output. Practice your conversions!
Yes indeed. Look all that stuff over. There are lots of things I can pick from in this. ALSO, make sure you understand the reasoning and math behind the Solar Energy Assessment (break even lab). We will be discussing this next class.
6. A 2m by 1m solar panel that is 20% efficient is producing 137 Watts of energy from sunlight. Determine the relative angle between the face and the sun.
Given what I have here:
