Hot answers tagged

46

No such liquid, safe or otherwise, can exist. Evaporation is a strictly endothermic process in all cases. The change in state from liquid to gas is marked by the individual particles gaining enough translational kinetic energy to overcome the mutual attractions present in the liquid phase to "fly free" in the gas phase. It is logically inconsistent for a ...


39

According to the Intergovernmental Panel on Climate Change (IPCC): "Greenhouse gases are those that absorb and emit infrared radiation in the wavelength range emitted by Earth." In order for a molecule to absorb and emit in the infrared (IR) region, its chemical bonds must rotate and vibrate in a manner that affects something called the molecule's ...


37

Thermite is a solid-solid reaction that I think would be greatly inhibited by any sort of intermediate. In either case, the smoke, the dust and the copious amounts of nitrous oxides produced will not make your asthma any better off than if you were to pick up a shovel. And then there will likely be other safety-, legal-, environment- and relationship-with-...


32

The wick temperature does not have to be the same as the flame temperature.The flame is hottest at the bottom, but the wick is hottest at the top. For a candle, the wick burning isn't the intended purpose of the wick; light comes from burning wax (more generally: fuel), you want to burn the wax not the wick. Rather the purpose of a wick is to help fuel ...


30

When you add salt to an ice cube, you end up with an ice cube whose temperature is above its melting point. This ice cube will do what any ice cube above its melting point will do: it will melt. As it melts, it cools down, since energy is being used to break bonds in the solid state. (Note that the above point can be confusing if you're new to thinking ...


28

Because fire is not the same thing as light. Michael Faraday did a wonderful job of explaining how the candle works, and I direct you to look at it (there are also Youtube videos giving a modern take on this work) if you're interested. In short, the candle produces light, not because it is hot, but because it is sooty. The particles of soot glow when they ...


27

Heating water on a hot plate is safe, because the hottest point is at the bottom of the pot. A lot of relatively small bubbles appear there without much overheating of the water, because there is a lot of nucleation at the uneven phase boundary steel-water. In a microwave, the hottest place is IN the water. The glass does not get heated by microwave (at ...


26

Diamond is one of the best thermal conductors known, in fact diamond is a better thermal conductor than many metals (thermal conductivity (W/m-K): aluminum=237, copper=401, diamond=895). The carbon atoms in diamond are $\ce{sp^3}$ hybridized and every carbon is bonded to 4 other carbon atoms located at the vertices of a tetrahedron. Hence the bonding in ...


26

Well, let's do some math: Assuming 30 mL of water is 30 g, and we want to heat our water from 20 °C to 90 °C, the energy required is: $$\begin{align}E&=C_Pm\Delta K \\ &=\left(4.18 \mathrm{\frac{ J}{gK}}\right)(30\mathrm{\ g})(70\mathrm{\ K})\\ &=8.778\mathrm{\ kJ}\end{align}$$ So how much power do we need to do this in a given time? "Instant" ...


24

I'll start by mentioning that there's no such thing as an exact measurement—there is always some measurement error. The only observations that can be numerically exact are counted numbers of discrete objects (e.g., the number of electrons in a neutral carbon atom is exactly 6). And I say "can be", because if the numbers are sufficient large, even ...


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. ...


22

In addition to the points Stian raised, it's important to realize that thermite is very difficult to ignite. The ignition temperature is very, very high, higher than you can easily get by burning ordinary fuels such as butane, which means that whatever ignition method you use, if it's hot enough to ignite the thermite, would most likely set off the butane ...


21

Let’s consider the following cases: getting $1\,\mathrm{mol}$ of $100\,\mathrm{°C}$ water on one’s skin getting $1\,\mathrm{mol}$ of $100\,\mathrm{°C}$ air on one’s skin getting $1\,\mathrm{mol}$ of $100\,\mathrm{°C}$ water vapour on one’s skin With the slightly irrealistic assumption that all of these liberate all their thermal energy to the skin while ...


20

Why won't water freeze if you put ice in it? It will, even at room temperature. You just need a big enough, cold enough ice cube. Don't believe it? Add a few drops of water to an ice cube in an ice cube tray (which is the same as adding an ice cube to a few drop of water). Wait a few seconds, turn the tray upside down. No water will fall, presumably ...


20

Many organic reactions are unreasonably slow and can take an extended period of time to achieve any noticeable effect so heating is often used to increase the rate of reaction. However, many organic compounds have low boiling points and will vaporise upon exposure to such high heat, preventing the reaction from proceeding in full. To address this, heating ...


20

When heat is leaving earth it leaves as infrared radiation. Greenhouse gases are gases that are able to absorb this infrared radiation. If we look at the infrared emission spectrum from Earth[1]: We can see that between $\pu{400 cm-1}$ and $\pu{700 cm-1}$, a lot of the infrared radiation is absorbed by $\ce{CO2}$. Gases like $\ce{N2}$ and $\ce{O2}$ don't ...


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 ...


19

A container that is not at ambient temperature will generate air currents around it. If you place such a non-ambient container in a balance, air currents will develop around the container as it heats or cools to ambient. These air currents will cause the balance to read incorrectly.


19

Your approach is quite correct, but as Jan already pointed out, it is incomplete. What you calculated is the difference in the electronic energy of the reaction $$\ce{H2 (g) + 1/2 O2 (g,\,{}^1\Delta_{g}) -> H2O (g)}.$$ Let's turn this into some kind of a tutorial and a little exercise, that you can try to reproduce at home. For a more detailed ...


19

Absolutely yes. Lighting a torch in such an environment would simply be the reverse physical process (and same chemical process) of what is done in our oxygen-containing atmosphere. In the chamber or alien world of hydrogen gas, providing an ignition source to a stream of oxygen would give a flame. The chemical reaction would actually be the same as if ...


17

Ron's answer is great, but I'd just like to touch on the mechanisms behind thermal conductivity so we can rationalize the differences between the behaviour of diamond, graphite, and metals: There are two ways in which heat is transmitted through solids: phonons and electronic conductivity. The latter occurs in electrically conductive solids, where ...


17

Compounds A-D all have the same molecular formula, $\ce{C6H12}$. We can burn each compound and measure the heat given off (heat of combustion). Since they are isomers, they will each burn according to the same equation $$\ce{C6H12 + 9O2 -> 6CO2 + 6H2O + heat}$$ Any differences in the heat given off can be used to say that a compound is more stable (it ...


17

Water has hydrogen bonding in it. Hydrogen bonding is some kind of intermolecular force (a tutorial and the wikipedia page) that is usually seen in molecules that have $\ce{OH}$, $\ce{NH}$ or $\ce{FH}$ somewhere in their structure. How does it happen? Hydrogen atom is really small (atomic radius: About 37 pm) When it bonds with some very electronegative ...


17

If you need a quick source of moderate heat, why not use a single liquid? Sodium acetate trihydrate ($\ce{CH3COONa·3H2O}$), for example, releases heat energy on crystallization... and it takes very little to trigger that crystallization! Commercial heat packs contain a "clicker", which makes a sharp vibration. You can make your own heat pack fairly safely ...


16

What actually happens in real life depends on a lot of things. Factors like the shape of the pot can make a big difference to how much of that 2500 W actually goes into vaporizing the water and how much of it is lost to the surroundings. If we assume that all 2500 W is going into the water, it makes things a lot simpler. If the water is already at the ...


15

From the comments: What do you want to do with the hot water? Swim in it. I was thinking of thousands of liters That's an interesting idea, but unfortunately, I don't think adding chemicals to a pool in order to heat it is a good idea (especially yellowish chemicals). The water temperature would drop in a few hours, tops, and you'd have to be ...


15

The Enthalpy $H$ is defined as $H=U+PV$. Therefore, $$\Delta H=\Delta U + P\Delta V +V\Delta P$$ For an adiabatic process, $q=0$. therefore from the first law of thermodynamics, $$\Delta U = q +w =q-P\Delta V$$ $$\Delta U=w=-P\Delta V$$ Substituting this in the first equation you get, $$\Delta H=V\Delta P$$ If $\Delta P$ is zero during the process (...


15

Most heat capacities go through a maximum as the temperature increases. $C_V = \left( \frac{dU}{dT} \right)_V$ so a maximum in $C_V$ corresponds to a minimum in $\left( \frac{dT}{dU} \right)_V$, i.e. the point where the temperature changes very little as energy is being supplied to the system. At this point (most of) the energy is being used to excite ...


15

Preliminaries Consider $U = U(V,T, p)$. However, assuming that it is possible to write an equation of state of the form $p = f(V,T)$, I don't have to explicitly address the $p$ dependence of $U$, and I can write the following differential: $$\mathrm{d}U = \underbrace{\left ( \frac{\partial U}{\partial V} \right)_T}_{\pi_T} \mathrm{d}V + \underbrace{\...


Only top voted, non community-wiki answers of a minimum length are eligible