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65

There's two main misconceptions in your question that cause your confusion. First, pressure doesn't cause higher temperature. This misconception is probably a result of a massive oversimplification with relation to the ideal gas equation. The actual relation is "increasing the pressure of an ideal gas while volume remains constant increases the temperature ...


61

It's not so much the pressure, but rather compression that creates heat. Heat is a measure of increased kinetic energy as molecules are forced into a smaller space. Water is not very compressible, and water at the bottom of the ocean is not confined to a significantly smaller space under pressure. The kinetic energy of water molecules at the bottom of ...


20

Colder water is denser until it reaches a temperature a couple degrees above freezing, then it gets lighter again. So the water at the bottom is at the specific temperature where it is densest: any heating makes it rise. Any further cooling makes it rise. See Why does the ocean get colder at depth? This further points out that without ocean circulation ...


7

TL; DR The material with the greater effusivity will be more likely to burn you upon contact. Analysis of a simplified case First consider the case where your palm comes into contact constant temperature wall. Often, we can consider your palm as a semi-infinite solid. The requirement for the semi-infinite approximation is that $T(x \rightarrow \infty, t) = ...


6

It has been covered above that there is little scope for water to be heated by compression to begin with. Another aspect is that the water at the bottom of the ocean has been there for a considerable length of time. Hence, if it had heated up to any large extent when the oceans were formed, there has been ample time for the heat of compression to have ...


4

There is another factor that I feel the other answers have overlooked, because there is a similar analogy with air, and air is compressible. Specifically, why is air in valleys often colder than at the top of the hill when pressure heats things? In reality, there are two different dynamics at work. One is adiabatic compression, which as has been ...


3

Due to the cylindrycal shape of the lamp, pressure forces act compresing the glass. Since the compressive strength of glass is quite high, $\approx 10\,000$ bar, atmospheric pressure is not enough to break it.


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 ...


3

So far there is no definitive answer as to why time flows forward. While many scientists speculate what may the cause of this asymmetry in time none of the options has been accepted as the one answer. Often these ideas lack a complete understanding of the underlying more complex principles that we cannot yet understand but here is one of the most common ...


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.


2

The simple answer to this question is that the specific heat capacity of a hot solid body you touch really doesn't matter too much for whether you burn yourself; what is much more important is its heat conductivity. The reason is that the surface of a hot body with low heat conductivity will rapidly cool down when you touch it (the blood in your body ...


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

Phase changes depend not only on temperature but also pressure. Like any substance, if you get the right combo of pressure and temperature you can keep it at a solid, liquid or gas or if you're really good keep it at the triple point and get all three phases in equilibrium with each other. For C02, the liquid phase is very likely outside of most temperature ...


2

Your error in method 1 was assuming the U = U(T) and not U = U(T,A). But, you would have been much better off starting out directly with entropy S = S(T,A), so that $$dS=\frac{C_A}{T}dT+\left(\frac{\partial S}{\partial A}\right)_TdA$$Then from the Maxwell relationship, $$\left(\frac{\partial S}{\partial A}\right)_T=-2\left(\frac{\partial \sigma}{\partial ...


2

In the method 1, you expressed increase of energy during adiabatic process as $$ dU=C_A dT $$ where $C_A$ is presumably capacity when the area $A$ is constant: $$ C_A = \left(\frac{dU}{dT}\right)_A. $$ But $C_A$ is not the right factor to use for adiabatic process, because in this process $A$ is not constant, but $S$ is. The energy increase formula is ...


2

You can't get an exact number for this without some assumptions. However, you can develop a relationship. What we do know is that one full period corresponds to some amount of time (my niave guess would be 1 second but I believe I have heard faster clocks which might be half seconds). So we introduce some constant C which is the amount of time reported on ...


2

While a funny-looking coincidence, this is not a valid alternative expression for entropy in general, since the entropy of a probability distribution (which are what rigorously hides behind the strange word "macrostate") is more generally given by $$ S = - k_B \sum_i p_i\ln(p_i) \tag{1}$$ and becomes only $$S = k_B \ln(\Omega) \tag{2}$$ in the case of a ...


2

Let us start with an example, called Langevin paramagnetism, where the magnetic moment is described classically, as a vector in three dimensions. Calling $\vec\mu$ this moment, $\vec B$ the magnetic induction and $\theta$ the angle between $\vec\mu$ and $\vec B$. The probability density of the angle $\theta$ is $\rho(\theta)=\frac{1}{\mathcal Z}\exp(\beta\mu ...


2

The whole point of statistical physics is that we don't really care about what the microstate could be, we just know that there's a great collection of similar ones that give rise to the same macrostate. I believe what you're questionning about is closely related to the ergodic hypothesis


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, ...


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

Most things along those lines are just simply the central limit theorem / law of large numbers. For example, when you compress a gas in an insulated container using a piston, its temperature goes up. Why? Because the moving piston accelerates gas molecules that bounce off it. And why does the temperature always go up by the same amount? Because there are ...


1

Assume the power of the microwave oven to be $P$ and that the instructions for cake and custard lead to the same temperature ($T$) of both when they are heated separately, then: $t_1=\frac{m_1c_1(T-T_0)}{\epsilon P}$ and : $t_2=\frac{m_2c_2(T-T_0)}{\epsilon P}$ where in the indices $1$ and $2$ refer to cake and custard, $m$ the mass, $c$ the specific ...


1

Assume that each swing advances the second hand by one second on the dial. You have a pendulum which is supposed to have a period of exactly one second. If the pendulum has got longer in length it will have a longer period and so it will take longer for the pendulum to advance the second hand by one second on the dial. So the clock will run slow. Just ...


1

You are on the right track. For a mechanical pendulum, the relationship is linear. You don't need to know how many swings of the pendulum corresponds to how many seconds. If the pendulum is x% slower, it will report x% fewer seconds per day. Now since length goes as $\ell = \ell_0(1+\alpha \Delta T)$ and period of pendulum as $$T = ...


1

I have an old pendulum clock (probably over 100 years old) still in working order. I'm pretty sure that the hands advance linearly with the number of swings of the pendulum. I keep my house about 10 degrees Fahrenheit cooler in the winter than in the summer, but don't really notice a difference in its timekeeping. The bob is suspended by a wire made of ...


1

Yes, it is a sort of contradiction. More precisely, it says that you cannot model water's liquid-to-gas phase transition with an ideal gas model, and any attempt to do so will be fraught with contradictions. Now as to some misunderstandings that you may have: first, be careful about thinking of potential energy as infinite at infinite separation: that's ...


1

I think that what you're asking can be done. Start at state 1 ($T_h,V_1$) and let the gas expand from $V_1$ to $V_2$ at constant $T_h$. This determines $Q_h$. Then let the gas expand from $V_2$ to $V_3$ adiabatically and reversibly. This determines $T_c$. Then compress at constant $T_c$ from $V_3$ to $V_4$. This determines $Q_c$. $V_4$ has to be such ...


1

In ideal gas model, temperature is the measure of average kinetic energy of the gas molecules. In the kinetic theory of gases random motion is assumed before deriving anything. If by some means the gas particles are accelerated to a very high speed in one direction, KE certainly increased, can we say the gas becomes hotter? Do we need to ...


1

DanielSank's answer is 100% rigth about the temperature issue. The question deserves much more argumentation because the book you use, and many others, are plain wrong. I will use numbers and the laws of physics to show my point: it is not true that If the mass of an electron, the Planck constant, the speed of light, or the mass of a proton were even ...



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