Urgent! You like or are good at explaining thermodynamics? This is your question, you have it, flaunt it!?
109470
This ain't my forte, but here goes.\n\nAdiabatic expansion is like what my freezer and air conditioner use to cool things off.  They take a pressurized gas and let it expand, which drops its temperature.  Then they let this low-temperature stuff suck up the heat in my Ben & Jerry's or in my house.  If you've ever discharged a CO2 fire extinguisher or used a can of compressed air to clean your computer, the CO2 is so cold it condenses the water in the atmosphere or the air on your computer is very cold.\n\nIsothermal means the temperature stays the same.  So if your gas expanded and it didn't get colder, something else went on to provide the heat to keep the temperature the same.  \n\nMaking heat is work, in physics terminology, so the isothermal expansion required more work than the adiabatic expansion.
569322
Use the ideal gas law \n\nPV/T = constant for the final and initial states.
575202
heat is a form of energy.
791970
Use the Gay-Lussac gas law: V/T is constant. Note that T here is the temperature in Kelvin, not in Celsius. Luckily 27 C = 300 K
96040
The phrasing is wierd, shouldn't there be a from between flow and the?\n\nIf two substances have different _temperatures_, heat will flow [from] the hotter [substance] to the colder [substance].\n\nThis is because heat, at the molecular level, is related to how fast molecules are moving. When a substance whose molecules are moving faster comes in contact with an object that's moving slower, the faster molecules will collide with the slower molecules of the second object, slowing down the faster molecules while speeding up the slower ones. Thus, heat (average kinetic energy of a substance) is transferred from the hotter substance to the colder one.
799142
Hi Kitty,\nHere is what, I think, you need.\n\nExothermic, dH < 0, Entropy increases (dS > 0), Reaction is spontaneous and dG < 0\n\nExothermic, dH < 0, Entropy decreases (dS < 0), Reaction is spontaneous only at low temps, if |T dS| < |dH|\n\nEndothermic, dH > 0, Entropy increases (dS > 0), Reaction is spontaneous only  at high temps, if T dS > dH\n\nEndothermic, dH > 0, Entropy decreases (dS < 0), Reaction is not spontaneous, dG > 0
776432
I am a HVAC trainer in the USA, and as such I have to explain the difference between energy (heat is a form of energy) and temperature. For the most part, my students do not posses any college background, and so I do my best to explain this in layman terms. There are chapters written on this subject, and to do it justice here …. well here goes.\n\nHeat is a form of energy, as is electricity, natural & propane gas, coil, gasoline, etc. Temperature is a measurement of how densely heat energy is packed into a substance. The more heat energy there is in a substance, the more vigorous the atoms and molecules will move. If the motion is vigorous enough, the substance can change its state and assume new characteristics (ice is different from liquid water, which is different from steam). \n\nThe temperature of a given amount of substance will increase as heat energy is added to that substance, and temperature will decrease as heat energy is taken (provided the substance does not change state). If a substance is at a temperature where it will change state, examples: ice~water @ 32deg F., …. and water~steam @ 212deg F. (at one atmosphere pressure), than the addition or subtraction of heat energy will cause the substance to change state, and the temperature will not change until the substance has finished changing state  “Sensible heat” is a form of energy that causes a change in temperature, “latent heat” energy causes a change in state.\n\n”Specific heat value” is the amount of heat energy it takes to rise or lower the temperature of one pound of a substance one degree F.  For instance, if we take one pound of ice at 0deg F., and put 1/2 BTU into that pound of ice, the temperature of that pound of ice will go up one degree F, thus .5 is the specific heat value of ice. If we continue to put energy into that ice, the temperature will go up one degree F. for every 1/2 BTU of energy we put in until it reaches 32deg F., at which time the ice will be at a temperature where it will melt if additional energy is put into it. We will have to put an additional 144 BTU’s to that one pound of ice at 32deg F., to change it into liquid water at 32deg F. \n\nOnce we have liquid water, we will have to put 1 BTU into that pound of water to raise the temperature 1deg F., thus 1.00 is the specific heat value of liquid water (please note: the specific heat value of liquid water changes slightly as the water’s temperature changes, but for most calculations, the constant “1.00” can be used). \n\nWhen the water reaches 212deg F. at one atmosphere (pressure at sea level), the water will once again be at a temperature where it can change state. As heat energy is added the liquid water, it will turn into steam. The amount of heat needed to change one pound of liquid water at 212deg F. into one pound of steam at 212deg F. is ….. ready ….. here it comes ….. 970 BTU’s …. That deserves a “wow”. \n\nThe reason I made such a big deal out of the 970 BTU’s is because it’s important that HVAC technicians and mechanics understand the concept that it takes a lot more energy to change the state of a substance than it does to change the temperature one degree Fahrenheit (or for that matter, one degree Celsius).
294164
Thw formular to use here is Q = mcT where Q = quantity of heat, m = mass, c = specific heat capacity, and T is the change in the temperature, thus here Q = 2.0g*(30-25)*0.90 = 9J. Notice that I sed Joules as the unit.\n\n2.For this one, a law which i cant remember the name right now is nedeed. Here we have C(s)+O2(g)=CO2 Enthalphy -390KJ\nand also Mn(s)+O2(g)=MnO2(s) Enthalphy = -520KJ\nFor this law to be appplied, we have to write the second equation the other way to become MnO2= Mn+O2 Enthalphy = 520KJ. Notice that i changed the sign when i reversed the equation. Then O2 could be canceled out on the left side of the first equation and on the right side of the second equation. This was we are adding. Finally adding, we get MnO2(s)+C(s) = Mn + CO2 Enthalphy = -390KJ+520KJ = 130KJ.\nand thats the correct answer
1003847
The study of heat and how it behaves.
753535
Just use the formula\n\nE = m*Cp*(Delta T)\n\nE=energy\nm=mass\nCp= Specific Heat\nDelta T = change in temperature\n\nCp= 0.51998 J/(C*g)\n\nGood Luck with the test!
732831
As for me I think all is persuasive. If you start to count temperature from the point when its length = 50cm you better write: L=50*e^(kt) cm. \nNow (iii): the constant k is valid to be unchangeable in rather narrow range of temperatures, in real physics k depends on t. And after melting the whole business does not work at all. On the other hand the function L=50*e^(kt) says that if t<<0 L is expected to shrink to 0, ignoring the fact that t cant be lower than absolute 0.
896452
Not sure the level with which you are learning, so feel free to ignore the thermodynamics explanation.  If you just want to answer the question, select D.\n\nLike most test questions, there is not enough conditions specified and truly, we could choose several anwers, but only after making assumptions that less likely to be in the testers mind.  I will assume a high school or intro level college course because above that this should be challenged as too vague.\n\nUse logic to realize that answer choice C is not true. Evaluating the others we realize that A means only heat is given off when bonds form and B means heat is absorbed or added during the formation of bonds.  D means conditions exist in which either situation is possible and E means that final product is in a state that has a higher value for entropy.  This only holds true when both Temp and pressure of a system is held constant, but we will eliminate later as well.\n\nBasically a bond can occur when the system goes from a high energy state to a lower energy state.  Formation of bonds is essentially a chemical reaction.  The Gibbs Free Energy equation:\n\n (delta)G=(delta)H-T(delta)S\n\nT=temp of system in Kelvin\n(delta)H is the heat energy change\n(delta)S is the change in entropy (disorder)\n\ndefines whether a reaction will occur spontaneously or not.  (deltaG) must be negative for this to be true. So we will assume \nusing this equation and recognizing that (delta)H tells us exo or endothermic if it is positive or negative , examination of the possibilities of (delta)H and (delta)S,\n\nIf (delta)H is negative and (delta)S is positive the equation will result in a negative value for (delta)G which means a favorible condition exists for bonding to occur.  Therefore, the condition of (delta)H negative (energy leaves as heat - exothermic) is true.  We can thus eliminate answer choice B.\n\nIf the opposite is true and (delta)H is positive and (delta)S is negative, the result is (delta)G that is positive and therefore not spontaneous.  no reaction or bonding occurs.\n\nIf both (delta)H is positive and (delta)S is positive, only a large enough (delta)S will overcome the positive (delta)H in the equation.  So if both are positive and T(delta)S is greater than (delta)H the bonding will occur.  Therefore, when (delta)H is positive (endothermic) the reaction may occur as well.  We can thus eliminate choice A.\n\nWe have shown that C is a true statement because conditions can be met in which the reaction is both exo or endothermic.  However, please note that in both conditions (delta)S is also positive, indicating an increase in entropy.  For the most part any system will always trend toward an insrease in entropy.  \n\nThe question then arises, what happens when (delta)S is negative (decreased entropy) and (delta)H is negative.  For most situations T(delta)S will be large enough to create a positive (delta)G from our equation.  In very rare cases will (delta)H - T(delta)S result in a negative answer.  However, the equation shows this as a possibility when (delta)G= large negative (delta)H - T x small negative (delta)S => large negative # minus a smaller negative => larger negative + smaller # = negative.  (Hope you follow my train of thought).  Therefore in the right situation - (delta) can be negative and bonding/reaction can occur when entropy decreases.  Eliminate E and C is the correct answer.\n\nCareful what you ask for.  Wikipedia site looks pretty informative but very detailed and you have to use a few different pages.
281886
Heat would an obvious example. The laws of thermodynamics would be another.
1003064
Kelvin, in 1851, gave the statement for second law of thermodynamics as "It is impossible to produce work in the surroundings using a cyclic process connected to a single heat reservoir."\n\n\nIn a general sense, the Second Law says that the differences between systems in contact with each other tend to even out. Pressure differences, density differences, and particularly temperature differences, all tend to equalize if given the opportunity. This means that an isolated system will eventually come to have a uniform temperature. A thermodynamic engine is an engine that provides useful work from the difference in temperature of two bodies. Since any thermodynamic engine requires such a temperature difference, it follows that no useful work can be derived from an isolated system in equilibrium, there must always be energy fed from the outside. The Second Law is often invoked as the reason why perpetual motion machines cannot exist.
826242
rubbing 2 things together to create heat.\nRub yor hands together really fast and see what happens.
64205
I'm not sure I understand the question.  Add some detail?
827524
Easier compared to one with a higher heat capacity:\n\nH/(m*Cp) = (delta T), where m is the mass of the object, H is the heat that must be added (or removed) to raise (or lower) its temperature, delta T is its change in temperature, and Cp is its heat capacity. From the above formula, you can see that larger values of Cp result in small temp. changes, for the same heat amount.