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The trick is that you're not just moving air, you're also increasing your volume. While the air is in the cylinder, it is at a very high pressure. A given amount of air takes up a rather small amount of space. When you push air into the BCD, you let it expand, decreasing it to a lower pressure (equal to the pressure of the water, actually). Now that same ...


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Using the ideal gas law: $P(h) = P_0 + \rho_w g h$, ${V_o P_0} = {V(h) P(h)}$ which gives $V(h) = {V_o P_0}{1 \over P_0 + \rho_w g h}$.


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As you go deeper, the pressure increases, decreasing the volume of the gas in the bubble. In water, the pressure is appoximately $14.7(1+\frac d{33})$ psi where $d$ is the depth is measured in feet. The volume goes as the inverse of the pressure.


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There is no easy way to calculate this for liquids because the heat exchange will depend on whether there is any convection in the liquid or not. You can calculate the solution for the heat (conduction) equation for your geometry, but this may or may not give the right answer. The problem is a lot better defined for solids which can not convect. The ...


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Instead of just adding boiling water as @Gert did, lets drain cool water and replace it with boiling. This way we'll keep the total volume of water constant $V=300$ gal. We can determine what volume of water to add $v_\mathrm{add}$. $$(V-v_\mathrm{add})\cdot T_\mathrm{orig} + v_\mathrm{add}\cdot T_\mathrm{add} = V\cdot T_\mathrm{target}$$ $$v_\mathrm{add} ...


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It's not the question asked, but looking at the power requirements might give some insight. Raising the water temperature requires a specific amount of energy, and the time constraint gives a required power. $$P = \frac{m C T} { t}$$ $$P = \frac{300\text{gallon }(1000\text{kg/m^3})(4.186\text{J/g K}) 14\text{degF}}{1 \text{hour}}$$ $$P = ...


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A typical hot tub will be coming to an equilibrium based on some forcing term $F$ adding heat to the system plus some proportional response $\lambda$ which loses heat to the environment:$$\rho \frac{dT}{dt} = F - \lambda (T - T_0)$$This is a linear ODE whose equilibrium temperature is $T = T_0 + F/\lambda.$ To increase $T$ as fast as possible you should: ...


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Assuming no heat is lost to the environment the heat balance on adding some boiling water ($212\:\mathrm{F}$) is given by: $$m_{bath}cT_{bath}+m_{added}cT_b=(m_{bath}+m_{added})cT_f$$ where: $m_{bath}=300\:\mathrm{Gall}$ is the initial amount of water, $T_{bath}=32.2\:\mathrm{Celsius}$, $m_{added}$ the amount of boiling water added, ...


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Heat transfer can occur by conduction, by convection, and by radiation. If you consider conduction through the bulk of the cup, the rate of heat transfer is directly proportional to the temperature difference across the material of the cup. As your experiment holds all variables equal except the temperature difference, cup A will lose heat at a faster rate ...


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Yes, agitation will generally promote heat transfer and reduce heating times (although quantifying the effect is not easy). But the effect is not related to the bulk speed of the kettle. When the water is heated a diffuse (poorly defined) boundary layer is formed on the bottom of the vessel. This layer is at a temperature that is slightly higher than the ...


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Coherent motion does not add to the temperature; so you would have to shake it violently, with random motions. Consider sound in air - this is a coherent motion - when you can no longer make out the sounds in a closed room, the energy of the sound waves has been transformed into heat.


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Visit http://www.science.uwaterloo.ca/~cchieh/cact/applychem/hydration.html You can see that the enthalpy of hydration is a two step proccess solvation and reverse crystallization, the ΔH(hydr) is actually positive so you have to give energy just to dissolve the NaCl in water, in order to seperate the water from the NaCl you need to account for the enthalpy ...


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Full derivation of volumetric flow for a siphon: Bernoulli’s equation in the absence of friction: $p_1+\frac12 \rho v_1^2+\rho gz_1=p_2+\frac12 \rho v_2^2+\rho gz_2$ Head loss in the case of friction (Darcy – Weisbach): $\Delta p=f \frac{L}{D} \frac12 \rho v_2^2$, where $f$ is the Moody friction coefficient. So: $p_1+\frac12 \rho v_1^2+\rho ...


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The initial velocity of the water through the hose doesn’t affect the velocity of the water that the siphon quickly settles down to. After only a brief moment, if water were a perfect fluid, the steady-state velocity of the water in a siphon would be $$v=\sqrt{2 g h} ,$$ where $g$ is the acceleration due to gravity at the Earth's surface, and $h$ is the ...


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i.e. everything else being equal, will I get a faster siphon if I start the process with a powerful pump than with a weaker pump? Or does that become irrelevant once I disconnect the pump and the siphon starts flowing? Simply put, only three factors significantly affect flow rate: Difference in height between inlet and outlet of your siphon. Roughly, ...


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Your setup of the problem isn't correct. Let V be the volume of water in the container. Then the rate of accumulation of heat in the container is equal to the heat in minus the heat out. The correct equation for this is: $$V\rho C_p\frac{dT}{dt}=q\rho C_p(T_{NEW}-T)-kV\rho C_p(T-T_a)$$This assumes that the tank is well-mixed so that the exit temperature ...


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The spot that you injected the red ink will turn red first of course.


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They can sometimes have this shape as they stretch downward from a faucet, as seen in this video, however, surface tension usually pulls the droplet into a roughly-spherical package. The reasons that raindrops have "streaky" trajectories in photographs is not that the raindrops are distended by air resistance, but rather that the short exposure time of the ...


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The diagrammatic representation of rain is indeed incorrect. Raindrops do not resemble teardrops. This Youtube video by Minute Physics provides a really cool explaination of that: http://youtu.be/8lBvC7aFB40


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When I google "speed of sound water temperature" the first hit is http://www.engineeringtoolbox.com/sound-speed-water-d_598.html Which contains tables of speed with temperature. Interpolate, calculate, enjoy.


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It's absolutely true. The bubbles are formed because of the speed with which the stream of liquid hits the surface. When you pour along the side, you take advantage of two things: first, the liquid slows down (because of viscous drag against the wall) second, by giving the liquid a solid surface, you prevent the formation of Rayleigh instability which ...


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Very interesting question. As you wrote yourself in your Edit it is hard to describe water via the ideal gas model. You have to introduce at least two important improvements of your ideal gas: Dipole - Dipole - Interaction instead of no interaction. Let's call this pair potential $V_d$ and note that for two given molecules $V_d$ is not only distance but ...


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Warmer water doesn't sink at any temperatur unless it is an iceberg and the coldest water floats. Vibrating atoms tend to move to slower moving atoms evening out. Vortexes could from the temperature change in the pot or in the ocean only then may you see warmer water lower than cold water in a continuous flow.


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A cooled surface is your best bet. Pressure-related phenomena are also a valid choice (basically, anything that changes the properties of the air-vapour mixture into conditions above saturation is fine), or combination of both. A cooled surface must have high thermal conductivity to drain the excess latent heat that is emitted during condensation (metal is ...


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Okay, this is how I would do it, I would set up cathode and anode to measure the conductivity. I would then compare that to a buffer solution of known ph. If you can't find a buffer solution, see if there is a chart you can find of Hydrogen ion conductivity relationship to pH.(But then you may have to take in temperature of solutions). ...



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