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Energy is needed to convert water to steam. This is called the latent heat of vapourisation and for water it is 2.26MJ/kg. So to boil away 1kg (about a litre) of water at 100ºC the kettle would need to supply 2.26MJ. Assuming the kettle has a power of 1kW this would take 2260 seconds. Given the unexpected interest in this question let me expand a bit on ...

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Has Musk done his homework? With regard to the basic idea of using nuclear weapons to release CO2 and thereby warm Mars, no, he hasn't. I suspect this was either Bored Elon Musk speaking, or perhaps the Elon Musk who didn't quite deny being a super villain ( 1-900-MHA-HAHA Elon Musk?) in that interview with Colbert. CO2's enthalpy of sublimation is ...

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The more surface area, the more heat transfer. Ideally you'd use a single-molecule sheet, but that's impractical. Practically, using crushed ice is very simple and very effective. You could also freeze water inside drinking straws or on baking sheets to achieve high area-to-volume ratios.

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Since neither of the answers given so far really answers the question, here's my $0.02: between convection (the flow of water of various temperatures around the kettle), and the fact that the heating element is at the bottom, the water is at various temperatures at various parts of the kettle at any time. Usually, the hottest is at the bottom, if the kettle ... 8 Temperature is a measure of average kinetic energy. When you have a kettle of water at 100˚C, some of the water molecules will have more-than-average energy, and some will have less. The more-than-average molecules are the ones that will turn to steam, carrying off their energy and lowering the average (and thus the temperature) for the remaining water. ... 8 Any black body in space radiates and ends up very cold, might even crystallize. The law of increasing entropy holds for closed systems, in this case the whole system: "all the radiation that left the black body + the black body itself" microstates. In the sense that a black hole behaves as a black body the same holds true, it cannot be considered a ... 7 Generally, bodies can have same internal energies, but have different temperatures, and vice versa, have the same temperature but different internal energies. Consider for example ideal gasses, where the internal energy is given as a function of temperature and heat capacity:$U={C}_{V}T$. If we have a monoatomic gas consisting of N particles, its heat ... 6 Because the diode lit for both ways of connecting your finger and soldering tip to the diode, the soldering tip has an AC voltage on it, and one side of the AC is connected to your building ground (the floor your bare feet were touching). The solder tip is heated by a coil for which it sounds like has 120 VAC on it. There is a short through the insulation ... 5 If you want to see all water in a container immediately turn to steam, you need a transparent container that you can seal. Fill the container 50% with water and tightly seal it. Place the container on an open flame and let it heat up. While it is heating, walk far away and watch the container through binoculars from some distance (e.g., 50-100 m should do ... 4 A large number of positive ions without any negative charges cancelling out the charge is extremely difficult to achieve in practice, which is why no one ever talks about it. You certainly can't use PV=nRT. An important consideration you left out is where are the negative charges? I understand that they are not in the container, but they are somewhere, ... 3 Pressure inside the container: zero Effective pressure on the container: proportional to$\frac{N^2}{r^4}$(where$r$indicates the size of the container) The situation: There are$N$protons inside an otherwise empty container. Assumptions: The container does not interact with the protons except for completely stopping them from moving outside (essentially, ... 3 The Heisenberg uncertainty principle applies to dimensions commensurate to h_bar, i.e. at the level of particles ( atoms, molecules, elementary particles). Temperature is a classical observable appearing in thermodynamics as a variable, but when analyzed from the emergent statistical ensemble it is not a variable but a statistical average of the kinetic ... 3 Comments already hinted at this, but the simple answer is that when you wrote the average velocity as a function of temperature, you were using a non-relativistic approximation (which is valid for most "everyday" situations, but not at extreme temperatures). Furthermore, since the particles have a mean velocity$\bar u$, you can't simply set that equal to ... 3 Then the number of microstates corresponding to this macrostate should be 1 (since all the particles are identical). The particles may be identical (have the same intrinsic physical properties), but that does not in any way mean that they are not distinct. It is most natural to assume the particles are distinct (they can be placed to different places, ... 3 We define everything in physics because it's useful. In this case it's useful when a fluid flows in a steady state system adn you want to look and energy flows. If a fluid flows through a box, and has a change in specific enthalpy$\Delta h$, with a flow rate of$\dot m$, then the power transferred to the fluid is$\dot m \, \Delta h$Sometimes all you know ... 2 Aluminum Oxide is a ceramic and comes in bits and pieces, see the production process in the link. How will you make it into a crucible? if not by melting and pouring it into a form? The melting point is 2072C . Generally clays also have to be fired to become stable, enter a phase tightly bound that only melting can destroy, or we would not have clay ... 2 It doesn't. It's a material priority. Specific heat capacity$c$is the heat required to raise the temperature of one kilogram of the material by 1 degree: $$c=\frac{dq}{dT}/m$$ It is not a material constant, though, as it depends of the state of the material while heating. Materials at different temperatures, volumes, pressures, etc. have different$c$. ... 2 It is possible to change the pressure by increasing the temperature - Pressure-temperature law, also known as Amontons' Law of Pressure-Temperature. The basic idea is in the diagram below: The law states that: The pressure of a gas of fixed mass and fixed volume is directly proportional to the gas's absolute temperature. As the temperature ... 2 I enjoy listening to this phenomenon as I wait for my teabag to brew. What is happening is that parts of the the ceramic cup heat up and expand. Because its shape is constrained (assuming it doesn't break), this produces a stress in the material that increases its resonant frequency (like tightening a guitar string). When it gets to Middle C, my tea's ... 2 This is a well known effect, and most clearly happens with hot instant cocoa which has just been mixed. Stir the cocoa while tapping the spoon on the mug, and you'll hear the pitch of the tapped spoon go down. Now start tapping while not stirring; the pitch will gradually go up by an octave or more. But, if you stir again, the pitch will go back down again. ... 2 Entropy is not a force, no. It is a "chaos factor", if you will. The more entropy, the less structured a system is. A system will always move towards the state that is most probable. The most probable state (macro-state) will be the state with most micro-state configurations. Consider as an example four coins of heads H and tails T. Flip them and you can ... 2 Edit: Looks like Steeven beat me to it with a similar explanation, but I'll leave mine here for posterity. Entropy is just a property of a system (in a given state), in other words a state function. I hope that by elaborating on the statistical interpretation of entropy you will gain some intuitive notion of the meaning of entropy. Consider the following ... 2 No and it's exactly the second law of thermodynamics that applies (in the Clausius formulation). You answered your own question. If you want more detail, assume that the solar panels have the efficiency of a Carnot machine. What's the efficiency then? 0% because there is no temperature difference. While solar cells are not mechanical engines, they are ... 2 The rate of temperature is obviously just the rate of solvent loss due to evaporation multiplied by the latent heat of vapourisation. The trouble is that I doubt there is any way of estimating the rate of solvent loss without actually doing the experiment. It will be dependent on lots of parameters related to the geometry of your apparatus e.g. bubble size. ... 2 John Rennie's answer is correct +/- 1% or so. Sure, you can lift something up, doing work instead of emitting heat. But you equally often lift something down, converting its potential energy to heat that your body emits. Besides, most of the loads you lift are small compared to the energy "cost" of running the chemical factory that is your body. Or say ... 2 See e.g. https://en.wikipedia.org/wiki/Saha_ionization_equation. Typical ionization energy for gases in planetary atmospheres are 14eV. The temperature on the surface of Venus is around 750K, which corresponds to an energy of 0.064eV. The exponential in the Saha-Langmuir equation will therefor basically completely suppress the existence of ionized atoms at ... 2 Firstly, it would be better to use actual, accurately measured numbers than the human 'experience': the human body is a poor thermometer and the mind plays tricks on us. But that hot water droplets lose heat and thus cool down in cooler air is an established fact and a consequence of the laws of thermodynamics. Regarding your three first bullet points, ... 2 Fourier's law$\vec\jmath=-\kappa\vec\nabla T$implies $$\frac{dS}{dt}=\int d^3x \frac{\kappa}{T^2}(\vec\nabla T)^2 + \ldots$$ where$\ldots\$ is other sources of dissipation (viscosity etc.). This result requires a little effort, but it is explained in standard fluid dynamics text books (chapter V of Landau, for example). Then the second law of ...

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Not even remotely possible. You would have to overcome the Strong Nuclear Force to rip a nucleus apart, and that is the strongest of all the fundamental forces. Here's a graphic from this site:

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This is a really simple issue. It seems your question is about getting intuition, so I'll entirely talk about that. Consider the Earth's atmosphere. If you go far up enough (a few hundred miles), you'll reach regions where the temperature is over 3000 K. Yet rockets aren't exploding in flames when they get there. Why not? They don't because the atmosphere ...

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