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7

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 ...


3

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 ...


3

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 ...


2

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 ...


2

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


2

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 ...


2

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 ...


2

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, ...


2

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, ...


2

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 = ...


2

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.


1

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} ...


1

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 ...


1

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: ...


1

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 ...


1

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 ...


1

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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