# Tag Info

## Hot answers tagged heat

24

$\ce{C + O2}$ is awfully complicated, so let's just pretend you've asked this: In a single act of the reaction $\ce{H. + H .-> H2}$, how is momentum conserved? That's a legitimate concern all right. After all, we are taught that this reaction does happen instantly, once given a chance, and that's in fact true. Also, we know that it releases a lot of heat. ...

20

Unfortunately, the question as stated is thermodynamically impossible. Let's look at the proposed reaction: $$\ce{CO2(g) -> CO(g) + O(g)}$$ This reaction is simply a bond dissociation (specifically, a carbon-oxygen covalent double bond is broken). We can look up the enthalpy change associated with it. From a table of values on Wikipedia, we find in the ...

18

The answer is NO. The article you quote makes a completely unwarranted generalisation that "all flames are hollow". This is true of some flames but only because the fuel that is burning is only able to burn when mixed with oxygen from the air. In those flames, the flame is "hollow" because only in regions where air can mix with the fuel ...

14

The decisive factor is the plant origin of honey. The major components ( reportedly by some beekeeper books of my passed away father ) have ratio glucose : fructose about 1:1 to 1:1.7. Glucose reportedly crystallizes more readily, triggering fructose crystallization as well. So honey closer to 1:1 ratio has stronger tendency to crystalize.(**) Aside of ...

9

There's not much to do, as you nearly spotted the difference by yourself: $U$ is a state function, thus is path-independent. In other terms, one can estimate it at any moment only with an initial and a final value (and express it in terms of its variables only). But with $Q$ and $W$, it's different: these are path functions (or process functions) ...

8

You’re looking at bond dissociation energies. They, however, do not give a good picture. A better place to start looking is the standard enthalpy of formation. The linked Wikipedia article provides an extensive list of compounds but only two matter: $\displaystyle\Delta_\mathrm fH^0 (\ce{CO}) = \pu{-110.525 kJ/mol}$ $\displaystyle\Delta_\mathrm fH^0 (\ce{... 8 Following are the Statement from “On Terephthalic Acid and its derivatives” By Warren De la Rue and Hugo Muller February 7, 1861: On heating, terephthalic acid sublimes without previously fusing(melting). The sublimate, which is indistinctly crystalline, has the same composition and properties as the original acid, and therefore, unlike other bibasic acids, ... 7 How is it possible, and what does it imply, when we say that the sum of two inexact differentials is an exact differential? In the example at hand it means that we divide the ways we can change the energy of the system into exactly two bins. The first we call heat, and all the other ones we call work. Heat is an energy transfer that occurs when two entities ... 6 @user1055 answered the question pretty much. I am just going to add a complementary answer which has the abstract to the paper which @user1055 mentioned. The abstract of the paper mentions the various products of the decomposition reaction: The thermal decomposition behavior of terephthalic acid (TA) was investigated by thermogravimetry/differential thermal ... 6 Answer A should be qualified by adding the words "... at constant temperature and pressure." Also, answers B and D can both cause the internal energy of an ideal gas to increase if the process is not carried out isothermally. 4 When two isolated atoms collide the total energy and momentum must remain with the two atoms so both are conserved overall. In fact in a reaction such as$\ce{H\cdot + H\cdot <=> H2}$the hydrogen molecule only lasts for a few femtoseconds. This is because even though the bond is formed the atoms will still approach one another (total energy being ... 4 To get heat, you need to go the other way (add$O_2$to$CO$). In steel mills,$CO$is a waste byproduct of the blast furnaces that gets used as fuel. It gets burned in boilers and makes high pressure steam. The steam spins turbines and makes megawatts of electricity. It's not a great fuel compared to methane, but you can't beat the price. 4 Latent heat of vaporization/evaporation: It's the amount of heat required for liquid ---> gas phase change of a substance at it's Boiling point. Not exactly. It's the amount of heat required for liquid ---> gas phase change of a substance at the particular temperature, there is usually and implicitly the substance boiling point. But generally, it is ... 4 A1: There is no displacement of the wall, forces are in equilibrium with zero net force. There are initiated stronger vibrations on warmer wall side by energetic hot air molecules. And weaker vibrations on colder wall side by energy leaching cold water molecules . A2a: If the wall initial temperature was the cold water temperature, then yes. Otherwise no, as ... 4 Imagine that you propose to climb Mount Everest. From your desk you can compute exactly how much energy you will need to reach the top of the mountain starting from base camp. That would simply be the gravitational energy difference between the two locations. With that information at hand you load up on supplies - dividing the gravitational potential energy ... 3 You're really asking about the difference between a physical change and a chemical change.* In a physical change, the chemical identity of the substance doesn't change. In a chemical change, it does. Suppose you take cube of sucrose (table sugar) at room temperature. If you heat it up, it will become warmer. But it will still be, chemically, sucrose. It ... 3 Since we are really interested in the heat effects of a liquid at somewhat normal temperatures. I am going to disregard the extremes displayed by diatomic gaseous elements and look at liquids at stp. For years water was considered the champion, but I reasoned that the more massive heavy water would show an even greater resistance to the transfer of heat ... 3 Yes, it is possible to fully decompose cinnabar to elemental mercury and sulfur. What conditions need to be satisfied in the new method? Before reaching to decomposition temperature, cinnabar phase transitions to a different polymorph. Data from this paper1 indicate that α-HgS (trigonal) → β-HgS (cubic) phase transition occurs at 673 K and is completed at ... 3 why is dH=dE at constant volume? It isn't, and nowhere in the problem or answer is this implied. First of all some definitions. For a combustion reaction, the enthalpy change can be equated with the heat of combustion at constant pressure, whereas the internal energy is the heat of combustion at constant volume: $$\Delta U = q_V ~~~~~\text{constant volume} \... 2 Can we justify that "For sublimation of a solid at 1 atm \Delta U>0 at low temperature and \Delta U<0 at high temperature?" No. \Delta U>0, always, for sublimation, because of the energy needed to separate the atoms or molecules in changing from the solid to the gas phase. As for the enthalpy,$$H = U +PV \Rightarrow \Delta H = \... 2 What are these people doing differently than in that school incident? The plaster casting of the human body is a totally different process compared to the incident at school. At school, it was a liquid plaster of Paris paste ready for molding. Nowadays, plaster casting is done in medical purposes differently. First they put thin stockinette (e.g., Delta ... 2 You should not separate the species on the left and assign enthalpy variations to each of them. On Scenario 1, you say that 1 mol of methane reacts with 2 mols of oxygen, and the variation of enthalpy for this is −889 kJ (since it is 1 mol of methane). If you had 2 mols of methane, and 4 mols of oxygen,$\Delta$H would be 2*(-889) kJ, and so on. As you ... 2 If a body absorbs a quantity of heat$q$its temperature will normally rise by a value$\Delta T$. The average heat capacity over this temperature range is defined as$C_{av}\equiv q/\Delta T$. The instantaneous heat capacity at temperature$T$is$C\equiv dq/dT$. This definition is not exact enough, however, until the path of heating is specified. From the ... 2 Consider a gas which was subjected to two different processes: isochoric and isobaric. Isochoric: If a certain amount of heat$dq$was given to this gas, then this heat will be completely used up in raising it's temperature since$dw$is zero. So, $$dq_v = dU$$ Also $$dq_v = nC_v dT$$ where$C_v$is constant volume molar heat capacity. Therefore we can say ... 2 Fire or flame is a luminous chemical reaction in the gas phase, i.e, such a chemical reaction is generating heat + light. Not all flames are very luminous e.g., hydrogen oxygen flame is very very light blue...hard to see. Candle flames are yellow because of glowing carbon particles. The blue/purple color comes from small molecules formed by the decomposition ... 2 As stated above tea is made up of hundreds of different substances, add milk to it and it becomes a more complex mixture. Anyway, detergent formulation chemists and textile chemists study those questions that you are trying to ask. Similarly, the cloth used in making a curtain is also a chemically complex material. Keep in mind the molecular dimensions. ... 2 No. If means that to maintain the temperature of the products equal to the temperature of the reactants, you need to remove heat, and, form the first law, if you do this, then$\Delta E$is negative. However, if you don't remove the heat, the temperature of the products will be higher than those of the reactants, and$\Delta E$of the system will then be ... 2 I'm glad that you have got good concepts but the problem is you're unable to connect the discontinuous dots. See, in order to answer your first two questions I would like to bring attention towards the difference of microscopic energy (energy possesed by the molecules present inside the system) and macroscopic energy (energy you get to see in the system). As ... 2 First law of thermodynamics is : I have divided the answer into two parts for better understand of the concept. For change in internal energy.$Δ U$or$\frac{ nfR Δ T}{2}$=$q $or$c*m*\ Δ T$±$ W$or$PdV$. Now , for an isothermal process ,$Δ T = 0 \$ . Therefore , change in internal energy and heat transfer is always = 0. But in your answer of the Q , ...

2

Is it to be open or can be closed ? If the former, the major loses can be due evaporation, keep the opening as small as possible and safe. If there is a wide opening as beakers have, try to put aluminium foil cap freely sitting on the beaker top, if applicable. Or a "watches/clock glass", typically used to cover beakers. Additionally, try to wrap ...

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