Free Energy. Passive Solar Hot Water
A thermosiphon solar collector system provides truly free energy. Unlike its active solar counterpart, passive solar energy uses neither pumps nor external input power. Passive solar collectors need nothing but energy absorbed from the sun to heat the water as well as to move that heat through the system. Once installed, a thermosiphon system will provide perpetually free hot water. During sunny periods the water in your collector is heated and pumped via the forces of convection to your storage tank, heat exchanger or indirect water heater, free of charge.
The laws of thermodynamics dictate that certain conditions exist to allow heat to flow properly throughout the system. If these mandatory conditions exist, convection will naturally cause a thermosiphon to occur within the closed-loop system. These physical requirements are as follows:
The output of the collector must be above the level of the return. The hot water storage tank must be located above the collector. The output of the collector must be piped to the top of the storage tank or heat exchanger coil. The bottom of the tank or coil must return to the bottom of the collector. There must be a minimum of hydraulic resistance and heat traps.
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Once these criteria are met, the system can function properly. Initially, as solar heat is introduced to the system, the water in the collector tubes becomes less dense, resulting in convection flow upwards toward the outlet of the collector. The heated water rises naturally to the storage tank or heat exchanger. This heated water enters the top of the tank and the cooler water at the bottom of the tank, being more dense, descends down the return pipe to the bottom of the solar collector. Providing there is a minimum of hydraulic resistance, such as sharp bends in the tubing; and the piping doesn’t contain heat traps in the form of dips causing a negative pitch, then the heat will continue its cycle unimpeded.
As more people seek alternative sources of energy to reduce their heating bills or to be free of the grid completely, renewable sources of power such as solar are sought more frequently. The thermosiphon solar collector system satisfies this increasing demand by offering a method of heating potable water or providing a source of radiant heat that’s completely free to operate!
Remember to evaluate this particular blog post very cautiously, the case and the alternatives have some distinctions. Charles LaRose has thirty years experience as an electronics technician and engineer and years of experience in the fields of plumbing/heating, electrical, and other building trades. He also has a keen interest in renewable energy sources and organic gardening. For more information on alternative energy, vermicomposting, and organic gardening, please visit HTTP://WWW.MYMAINESOURCE.COM Article Source: http://EzineArticles.com/?expert=Charles_LaRose
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Chemistry question about gibbs free energy? Waht is the standard gibbs free energy of formation of water vapor at twenty-five degree celcius, if, for the reaction shown below under stadard conditions, delta H=-484kj/mol and delta s-89j/molK 2H2(g) + O2(g)–> 2H2O(g) A. -457KJ/MOL B. -395 C.-229 D. Water vapor doesn’t form at 25cecius. I choose A but is wrong….
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The equation to find the gibbs free energy is delta G=deltaH – T * delta S T is temp in K T=25+273=298 K So just plug in your numbers and watch you units!!! DeltaG=-484000 j/mol – 298 * eigthy-nine deltaG= 457,478 j/mol = -457 kj/mol but you’re not done yet The standard gibbs free energy is for the formation of only one mol your equation has 2 moles so divide by 2 -457/2=-229 kj/mol The answer is C.
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Gibbs free energy and entropy? According to my book, the answer to “If the chemical reaction A -> B + C has a positive G, increasing the temperature will…” Is: Because G = H – TS, increasing the temperature will make G more negative. Is this always the case, or can there be a negative entropy which would cause the reaction to become less likely to occur as temperature increases?
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There can be but not in the reaction you were given. Entropy = disorder (simplistically). In your reaction one molecule (A) turns into two molecules. Thus the disorder has increased and entropy is positive.
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There can be but not in the reaction you were given. Entropy = disorder (simplistically). In your reaction one molecule (A) turns into two molecules. Thus the disorder has increased and entropy is positive.
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You understand this, and it sounds like your book is wrong. You need to know the sign of S in order to be able to tell (to first order) whether a spontaneous reaction become “more spontaneous” or “less spontaneous” as the temperature increases. For example, consider the reaction at 270K (-3C) and atmospheric pressure: H2O(liquid) H2O(solid) (i.E, Water freezing to ice) This is a spontaneous reaction, but the S for this reaction is negative (a liquid is transforming to a solid, S in such cases is always negative). H, of course, is negative and equal to the negative of the enthalpy of fusion for water. If we increase the temperature above 0C, this reaction won’t be thermodynamically favored. In this case, increasing the temperature *decreases* the “spontaneity of the reaction because S is negative. On the other hand, the reverse reaction: H2O(solid) H2O(liquid) (i.E, Ice melting) is *not* spontaneous at -3C and atmospheric pressure. Now, the S for the reaction is positive and H is positive (= the enthalpy of fusion), but at this temperature -T*S is not sufficiently negative to overcome the positive value of H, so G is positive. If we increase the temperature above 0C, however, the reaction becomes spontaneous. Increasing the temperature increases the “spontaneity” of reactions with positive S. Note that this the explanation above is only approximate. Both the S and H of a reaction depend on temperature. It’s commonly assumed that for small temperature changes, these quantities can be treated as constant (independent of temperature). This is what I meant when I said “to first order”. Things can get a little more complicated when one deals with large temperature changes and the consequent changes in S and H can't be neglected.
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You understand this, and it sounds like your book is wrong. You need to know the sign of S in order to be able to tell (to first order) whether a spontaneous reaction become “more spontaneous” or “less spontaneous” as the temperature increases. For example, consider the reaction at 270K (-3C) and atmospheric pressure: H2O(liquid) H2O(solid) (i.E, Water freezing to ice) This is a spontaneous reaction, but the S for this reaction is negative (a liquid is transforming to a solid, S in such cases is always negative). H, of course, is negative and equal to the negative of the enthalpy of fusion for water. If we increase the temperature above 0C, this reaction won’t be thermodynamically favored. In this case, increasing the temperature *decreases* the “spontaneity of the reaction because S is negative. Never the less, the reverse reaction: H2O(solid) H2O(liquid) (i.E, Ice melting) is *not* spontaneous at -3C and atmospheric pressure. Last but not least, the S for the reaction is positive and H is positive (= the enthalpy of fusion), but at this temperature -T*S is not sufficiently negative to overcome the positive value of H, so G is positive. If we increase the temperature above 0C, however, the reaction becomes spontaneous. Increasing the temperature increases the “spontaneity” of reactions with positive S. Note that this the explanation above is only approximate. Both the S and H of a reaction depend on temperature. It’s commonly assumed that for small temperature changes, these quantities can be treated as constant (independent of temperature). This is what I meant when I said “to first order”. Things can get a little more complicated when one deals with large temperature changes and the consequent changes in S and H can't be neglected.
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How do I calculate reaction free energy? Im not sure how to calculate the free energy for this problem. I found this equation in my notes: delta G = -RT ln K, but Im not sure if its the correct one for this problem. Can someone please explain the steps I need to take to solve this? Thank you Calculate the reaction free energy of: 2SO2(g) + O2(g) —> @SO3(g) when the concentrations are 0.5MOL/L (SO2), 0.5MOL/L (O2), And 0.5MOL/L (SO3). The temperature is 700K and for this reaction Kc=45 at 700K. It would be great if someone supplied the steps to work this out as I’ve no idea where to start or how to use the equation I've found (if its even the right one)
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