# Tag Info

0

The derivation is not correct. The mass within the control volume is $\rho A\Delta x$. The rate of energy accumulation within the control volume is $\rho A\Delta xC\frac{\partial T}{\partial t}$. So the heat balance should be:$$\rho A\Delta xC\frac{\partial T}{\partial t}=Q_x-Q_{x+\Delta x}$$Dividing by $\Delta x$ and taking the limit as $\Delta x$ ...

0

First of all: you and the people in your course are most certainly not the only students encountering this problem. In my 2nd and 3rd university year I had it myself. The reason is that physicists use abusive notation. And they do it a lot. Mathematicians have less of a problem with these things. In this answer I will try to use intuitive terms and be ...

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 ...

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

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, $$\left(\frac{\partial S}{\partial A}\right)_T=-2\left(\frac{\partial \sigma}{\partial ... 0 Not sure about your question. Here are some suggestions: In thermodynamics we define internal energy as U = U(S, V) (number of particles is constant). Then we write it's differential:$$dU = \frac{\partial U}{\partial S}dS + \frac{\partial U}{\partial V} dV$$Now we recall 1st law of thermodynamics dU = \delta Q + dW  together with 2nd (or it's ... 0 After a long discussion with "curiousone" I would like to like to share the relevant points of our discussion (hopefully I will do them justice) and some extra bits I added after thinking it over First Law of thermodynamics While the equation$$TdS = k_b T \ln N dN + dU -PdV $$is quite general to any system where particle number is not conserved. We ... 0 Every proportional relation , even with a coeff 1 , includes potentially a constant. Obviously, changing its value for anything else , changes all the physics. It remains to demonstrate that adjusting the units to maximize the number of coefficients 1 has a physical meaning. Elegance is a strong indication of relevance but not a proof. Ask Susy. However, ... 0 Instead of just adding boiling water as @Gert did, lets drain cool water and replace it with boiling. This way we'll keep the total volume of water constant V=300 gal. We can determine what volume of water to add v_\mathrm{add}.$$(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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These answers are way too complicated. It's not rocket science. When a charged particle moves it creates electromagnetic radiation. Matter is made of charged particles. Since everything not at absolute zero is moving around and everything is made of charged particles you're gonna get a spectrum of EM radiation coming off matter. If the thing is ...

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 ...

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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. $$P = \frac{m C T} { t}$$ $$P = \frac{300\text{gallon }(1000\text{kg/m^3})(4.186\text{J/g K}) 14\text{degF}}{1 \text{hour}}$$ $$P = ... 2 The simple answer to this question is that the specific heat capacity of a hot solid body you touch really doesn't matter too much for whether you burn yourself; what is much more important is its heat conductivity. The reason is that the surface of a hot body with low heat conductivity will rapidly cool down when you touch it (the blood in your body ... 1 A typical hot tub will be coming to an equilibrium based on some forcing term F adding heat to the system plus some proportional response \lambda which loses heat to the environment:$$\rho \frac{dT}{dt} = F - \lambda (T - T_0)$$This is a linear ODE whose equilibrium temperature is T = T_0 + F/\lambda. To increase T as fast as possible you should: ... 2 Assuming no heat is lost to the environment the heat balance on adding some boiling water (212\:\mathrm{F}) is given by:$$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 The maximum potential energy is limited by the energy content released by the one liter of boiling water. you are able to boil away all of the water - this gives you the total energy released. The obvious answer is to follow Watt and build a steam engine - the height of the piston is the measure of the potential energy extracted, U=mgh. You can use a ... 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) = ... 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 ... 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 ... 1 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 According to e.g. https://en.wikipedia.org/wiki/Greenhouse_gas#/media/File:Atmospheric_Transmission.png, it is clear that the term Infrared radiation, spanning from λ=800 nm to over λ=50000 nm, is misleadingly broad. Instead, I propose to speak in terms of solar radiation (UV+optical+near-infrared between 300 and 2000 nm) and thermal radiation from earth ... 56 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 ... 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 ... -1 In any process total entropy can not decrease. It's s second law of thermodynamics. But still why is it so? I can not give a good explanation. Any system consists of atoms, photons, etc., these particles move around, collide. Motion of each of them is described by laws of mechanics. It turns out that laws of mechanics have a very interesting property: ... 0 If the engine is operating in a cycle, the change in entropy must be zero, since entropy is a function of state. However, the heat taken in from the hot reservoir divided by the temperature of the hot reservoir must be less than the heat released to the cold reservoir divided by the temperature of the cold reservoir. This means that the net entropy ... 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 ... 3 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 ... 2 Let us start with an example, called Langevin paramagnetism, where the magnetic momentum is described classically, as a vector in three dimensions. Calling \vec\mu this momentum, \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 ... 0 The main reason sum of countable set instead of integral over continuous volume is introduced in statistical physics in the context of classical physics is simplification of mathematics needed to make use of the probability theory to explain basics of information theory and derive the very notion of information entropy, given by the p\ln p formula. Or use ... 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 ... 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 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 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

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 ...

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 ...

0

If you've ever scuba dived you'll know that the pressure in a tank falls as it cools down to ambient after filling, and you're advised never to leave the filled tank in a hot car. You can use the ideal gas law from thermodynamics to calculate the proportionality of pressure and absolute temperature in the circumstance of constant volume and constant number ...

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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 ...

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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 ...

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 ...

0

Let's assume that the initial temperature of the sphere is not 1000C, so that, at least, it is not above the critical temperature of water. Let the initial temperature be $T_0$ and let $T_{\infty}$. Let's also assume that the pressure is high enough so that the water close to the sphere does not boil. My goal will be to arrive at an upper bound to the ...

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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 ...

0

Newton's Law of Cooling actually states: $$\frac{dQ}{dt}=-kA[T(t)-T_a],$$ where $\frac{dQ}{dt}$ is the heat flux, $k$ the heat transfer coefficient, $A$ the sphere's surface area, $T(t)$ the sphere's temperature as a function of time and $T_a$ the ambient temperature (assumed constant). When the sphere drops in temperature by $dT(t)$ then it loses an ...

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 ...

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.

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

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 ...

0

The following answer assumes that you have defined the thermodynamic entropy function $S$, with the relation $dS=\frac{dQ}{T}$. Recall that the first law in differential form reads $dU=dW+dQ=dW+TdS$. For a cyclic process $\gamma$, we require $\oint_\gamma dS=0$. Hence, $\oint_\gamma dS=\oint_\gamma \frac{dQ}{T}=0$. Now consider what happens if $\gamma$ is a ...

0

We don't need to go to astrophysics for this. In the reversible expansion of a plain vanilla ideal gas, if one does not add sufficient heat, the temperature will drop (and, by this definition, the heat capacity will be negative). This can happen any time work is done such that there is not enough heat added to increase the internal energy. This is why ...

0

There are certainly systems that have negative heat capacities, and in fact they come up all the time in astrophysics. As a general rule, gravitationally bound systems have negative heat capacities. This is because in equilibrium (and remember we can't do classical thermodynamics without equilibrium anyway), some form of the virial theorem will apply. If ...

2

Check out this question here: Why does a candle blow out when we blow on it? Our breath is 16% oxygen and only 4% CO2 Basically when you shake the match violently you move the fire away from its source of fuel. The flames themselves will have some momentum which you can see from the way the flame "tilts" as you move the match. When you move the match very ...

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