# Tag Info

40

You blow away the flame from its fuel source. If you would blow less hard the flame might burn harder because more air is supplied to the flame (similar to a Bunsen burner). Because normally the flame of a candle gets its oxygen through a convectional airflow generated by the heat of the flame. The reason why the flame is blown away from the candle is ...

17

Combustion is a gas phase reaction. There are two requirements to generate a stable flame. Firstly the temperature must be high enough to vapourise the combustible material (wax in this case), and secondly the temperature must be high enough to generate the activation energy needed for the reaction. Heat is needed because gas phase molecules of wax and ...

7

If you look closely at the candle flame, you will notice that the flame hovers just over the wick, but does not touch it. This is because the flame is boiling the wax, which becomes a vapor, which then burns. All of these processes are driven by the heat from the flame. As you blow on the flame, you moving it away from the wax and disrupt this process. ...

3

Negative temperatures can only exist in a system where there are a limited number of energy states. Then as the temperature is increased, particles move into higher and higher energy states, and as the temperature increases, the number of particles in the lower energy states and in the higher energy states approaches equality. (This is a consequence of the ...

3

If you rearrange your equation to solve for $Y$, $$Y=(X+125)\frac{-30+70}{360+125}-70$$ This reduces to $$Y=\frac{8X-5790}{97}=0.08247X+59.6907\approx0.08X-60$$ which is about what the textbook obtains. The difference in values you are getting is completely due to the approximation that the solution manual uses. If you use the exact values, as you did, ...

3

As fibonatic noted, you are blowing the flame activity away from the wick, but that's not the entire story: if the wick were still the same temperature it would immediately reignite. You are super-cooling the system by introducing a large mass which can't be heated enough to sustain the fire in time. This is, incidentally, one of the primary reasons why ...

2

If You know density $rho_r$ at some temperature $T_r$, there is a following formula for density: $rho=rho_r[1+b(T-T_r)]$, where $rho$ is the density at temperature $T$ and $b$ is called coefficient of cubical expansion, evaluated at reference temperature and density ($rho_r$ and $T_r$). It is valid for liquids.

2

I am not an aerodynamics specialist, so the following is almost certainly a huge oversimplification (or maybe downright wrong), but I think it might help with intuition. Suppose you have an amount of energy $E$ available to spend, and you are trying to accelerate an object of mass $M$. Suppose you can impart the energy in the form of kinetic energy to an ...

2

The pressures are moderate so ideal gas is a good assumption. Moreover the temperatures equalize quickly so you can treat your problem isothermally with regard to the end states: $pv=RT$ gas equation $T_1=T_2$ so $p_1v_1=p_2v_2$ or $v_1/v_2=p_2/p_1$ If I take your pressure reading between 0 and 5s, I estimate it to be 92kPa. For the compressed state about ...

2

The difference between solid and liquid lies in the atomic structure. Ice is crystalline (and therefore in an ordered state) while water has no such ordering. It is amorphous. So the reason for the abrupt change in state is that something cannot be ordered and unorderd at the same time. Now you may say "Hey, why not have some regions that are orderes and ...

2

Heat source (or heat reservoir) and heat sink are terms used to describe thermodynamic cycles. It is easy to recognize them when You think about closed cycles. But vehicle engines are usually internal combustion type engines and as such perform open cycles. "Internal combustion engines are primarily heat engines" and "heat engines are often confused with ...

2

The book is correctly saying that entropy (S) of the system only depends on the state of system (Pressure, Temperature, Volume; P,V,T). The entropy can change, but if it does there is a change in state (P,V and/or T). However, if the change in entropy is being expressed as a function of heat, which is not a state fuction, then it matters whether the ...

1

I think that the simplest way to wrap one's head around negative temperatures is that one: $$\beta = 1/T$$ The point is -- it is much more "physical" to describe a temperature of a body in terms of $\beta$. (We are using inverse of $\beta$ for a number of practical and historical reasons, but nevertheless.) The larger that quantity $\beta$ -- the lower the ...

1

A candle flame points upwards because flame is extremely hot, and thus less dense than air (by a routine approximation using the Ideal Gas Law), and thus rises. A hot air ballon floats for the exact same reason. As mentioned in the "possible duplicate" question, this effect disappears in the absence of gravity.

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The direction of fire is whatever way the wind is blowing. Seriously. In a calm location, the heated air goes upwards. As to the composition: it's a bit tricky. The simple bit is that various molecules produced during combustion (usually oxidation) are very energetic, and shortly after combustion, they cool off by emitting photons. If you look closely, ...

1

I am not aware of a better name for this sort of point in the scientific literature. It doesn't exactly shock me if you refer it as an eutectic point, but you'd have to explain it in its first occurrence in the text. Otherwise, just assign a letter to it (just avoid “lambda point”!), and use that throughout your paper.

1

(Full disclosure, I didn't RTFA, and I don't have time to.) Just to review, $f\left(x,v,t\right)dxdv$ is the number of particles with positions between $x$ and $x+dx$ and velocities between $v$ and $v+dv$. First, why is there no integral over position? In principle, there should be. However, assuming that $d$ is small, $f\left(x,v,t\right)\approx ... 1$pV = NR_mT$gas equation$pv=R_mT/M=p/\rho$gas equation using specific volume$pv^n=const$polytropic$M$is the molar mass and$R_m$the universal gas constant$n=\kappa=c_p/c_v$for the isentropic case (loss of rigidity is assumed to mean frictionless here). We know that the change in volume for both volumes as well as the pressure after the change ... 1 Work and heat are not state functions. In this problem, there is no heat or work done by the system as a whole (considering both parts of the container), due to insulation and no change in total volume. The work done on one part equals the work done by the other part. There is no heat by either part due to insulation. This is an adiabatic process. There is ... 1 I think it is important to not loose yourself in calculations. The method in your book probably starts from saying that considering an object$\delta Q$that has the following general form:$\delta Q = C_vdT + hdV$say, then it is an exact differential iif$\left(\frac{\partial C_v}{\partial V}\right)_T = \left(\frac{\partial h}{\partial T}\right)_V$... 1 Your kettle needs incoming energy from the heating element to turn water in to steam. Steam bubbles forming and collapsing make the familiar sound. Early on many of the steam bubbles don't make it to the top because they cool off when they rise away from the heating element. This is why the familiar rumbling sound starts way before the water boils. The ... 1 Your own answer is along the right lines: the key is that this is not a quasi-static process, and the system goes out of equilibrium. However, in practice the piston probably won't move so rapidly that you can't define the pressure at all. Instead, the system will lose energy equal to$W_2\$ (to a good approximation), and the environment will gain energy ...

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