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Simple answer is no!! Blankets only work in air. Since air is a bad conductor of heat and blacnkets use this air to provide as a cushion for your heat and the external environment. Since in water there wont be much air and water too is a bad conductor of heat. Hence you should not use blanket in water


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There's a tiny trace of helium in air, about 5 parts in a million. The pressure isn't high enough for it to freeze no matter how cold it gets


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Here's another prospective. The temperature o an object (particle) is a function of its energy. Theoretically there is no limit to the energy we can keep adding into a system. However objects emit radiation that is dependent on their temperature. Object with higher temperature emit radiation with shorter wavelength. According to quantum mechanics the ...


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I believe the confusion is that you believe pressure will always increase as temperature increases. This is only the case in a closed environment such as inside the tire. In an environment such as the atmosphere which is, essentially, in an unconfined environment, the density will decrease with temperature as well. This does not happen inside of a closed ...


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The lens works because it takes all the sunlight falling on its area, $A_1$, and focuses it onto a small spot $A_2$. The intensity in the spot is the intensity of the sunlight multiplied by $A_1/A_2$. Exactly the same applies to a mirror. So provided your mirror has the same cross sectional area as the lens, and provided it can focus the light as ...


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The rise in temperature would not be linear. As you are heating a piece of material, the energy loss due to radiation will increase (and so will the energy loss due to conduction and convection in air, but we are not going to discuss those here). Radiation losses can be estimated by a simple formula for black body radiators called the Stefan-Boltzmann law: ...


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Because what you are doing is a flow process, with mass inflow and no mass outflow, you need to use the thermodynamic equation: $dU_{cv}={m_{in}d}{H}_{in}-{m_{out}d}H_{out}+\delta Q-\delta W_{shaft}$ If you insulate your air cylinder well enough, $\delta Q = 0$. Assuming that your air cylinder does not deform, $\delta W = 0$. Since you are filling your ...


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The energy change a gas is expressed by: $$dE = TdS - pdV +\mu dN$$ Admittedly, the process of filling a cylinder is rather complicated because several variables in this equation vary, and the way you fill it will influence their behaviour, but you can see that the result will give a production of heat. When you fill the cylinder you are connecting it to ...


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Use this equation: $$\frac {P_1V_1}{T_1}=\frac {P_2V_2}{T_2}$$ Before, the pressure is 1 atm, the temperature is about 20°C and the volume is many cubic meters. After, the pressure is above 100 atm, the volume is about 1 cubit meter, and the temperature is significantly warmer, say 100°C. Filling in the equation: $$\frac {(1atm)V_1}{393K}=\frac ...


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I would explain it by simply using the first law of thermodynamics: $$Q-W=\Delta U$$ $Q$ is heat added to the system, $W$ is work done by the system, $U$ is the internal energy. Keep in mind that internal energy is closely bound to temperature, so a change $\Delta U$ also results in a temperature change (which is what we are talking about). If you look at ...


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There are, by my count, 33 other official weather observation stations within Sydney. Here's a map: With regard to the Sydney - Observation Hills (Station ID 066062) station -- That freeway didn't exist when the station was built. How could it? That station dates back to 1858. The station was moved 150 meters to its current location in 1917. That tiny ...


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It no doubt does. Solar radiation will heat surfaces, and the rate at which a black surface absorbs that heat is greater than for other colors. That heat is then partly lost by radiation, partly by conduction, and partly by convection. If you assume the last two terms will be the same regardless of the color, then you can see why a black surface will become ...


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Yes, the probability for the electron to be found inside the nucleus is, for some atomic orbitals, non-zero. However, you must recall that these orbitals usually assume a point charge for the nucleus, and so they may not be a valid when you "zoom in" to the nucleus. Nevertheless, there's nothing inherently wrong with the electron being where the proton is - ...


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This is a very old question, but none of the answers fully address the question. I'll frame my answer in terms of answers to a series of questions: How much does temperature vary with altitude? Why does pressure vary with altitude? Why does temperature vary with altitude? What about the second law of thermodynamics? Why is the Tibetan Plateau so cold so ...


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Since gas molecules are affected by gravity, wouldn't that make gas molecules at higher than average elevation slower (at the top of their ballistic parabola) and thus colder than air molecules accelerating to the ground? In non-relativistic theory no, because in thermodynamic equilibrium temperature has to be the same everywhere. The slowing down does ...


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AdS black holes exist in various dimensions, $p=3$ is not the only choice. The parameter can take on values above or below $3$. One famous example is the three-dimensional BTZ black hole, and higher dimensional ones are also frequently used in the correspondence. Furthermore, I think there is a misunderstanding on the concept of a "gravity dual". The metric ...


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It seems that most places I've read (on the web) people refer to the depressurization of water (in a vacuum or space) as "boiling" but I have very rarely seen this referred to as an out-gassing of the water's internal dissolved gases (nitrogen, oxygen, CO2, or any other gases that may be used in a space vehicle). However, unlike traditional boiling which ...



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