# Tag Info

## Hot answers tagged temperature

6

The important part is that it works "between two temperatures whose difference is 100°C". Celsius and Kelvin are not the same, but their degrees measure the same. You can see that by direct substitution $T(ºC)=T(K)+273.15$ therefore $T_2(ºC)-T_1(ºC)=T_2(K)+273.15 -(T_1(K)+273.15)=T_2(K)-T_1(K)$

6

if the pressure is the same, there in no net force on the piston, so it will remain at rest, so neither gas will expand, and this will stay this way because there is no heat exchange that could change the pressure on either side. Ask your teacher to return his diploma.

4

A difference in degrees Celsius is equal to one in degrees Kelvin. So 150 K - 100 K = 50 K = 50 C = 423.15 C - 373.15 C. The absolute values cannot be compared without taking the 273.15 offset into account.

3

The amount of energy liberated per gram of material per second in the fusion reactions depends on the density, the mass fraction (hydrogen, $X$, helium, $Y$, and all others $Z$) and temperature: $$\epsilon = \epsilon(\rho,X, Y, Z, T)$$ Typically we express the energy generation rate as a power law, $$\epsilon\propto\rho^\alpha T^\delta.$$ though the ...

2

The relativistic doppler shift of frequency is given by $$f_o = \frac{f_ s}{\gamma (1 + (v/c) \cos \theta)},$$ where $f_o$ is the observed frequency, $f_s$ is the emitted (source) frequency, $\gamma = (1 - v^2/c^2)^{-1/2}$ and $\theta$ is as you define. This is a standard result - e.g. ...

2

Your teacher is wrong... in a way. What he probably meant was same mass, different temperture.

2

First, note that the equation you use is only valid when all relativistic particles are in thermal equilibrium. The more general equation, which allows for particles with different temperatures, is $$g(T) = \sum_B g_B\left(\frac{T_B}{T}\right)^4 + \frac{7}{8}\sum_F g_F\left(\frac{T_F}{T}\right)^4$$ where $T$ is the photon temperature and $T_B$, $T_F$ are ...

2

Your post seems an awful lot like a homework question, so you should probably tag it as a homework-like question. If by "cup" you mean 250 mL on the dot, then we can say that the ratio of boiling water (100 Celsius) to room temperature water (22 Celsius) to achieve 80 degrees is $$\frac {(T_1)(V_1) + (T_2)(V_2)}{V_1 + V_2}$$ In your case, we have: $$... 2 For viscous hypersonic flows, the heating takes a form:$$ q_w = \rho_\infty^N V_\infty^M C  where the parameters $N$, $M$, and $C$ depend on the configuration and $q_w$ is the heating in $W/cm^2$ (this is all from Hypersonic and High Temperature Gas Dynamics and I highly recommend this book). For the stagnation point (like the leading edge of a body): ...

2

It's dangerous because the Leidenfrost effect is only temporary; it only works as long as the temperature of the skin exposed to liquid nitrogen is much higher than the boiling point of liquid nitrogen. Obviously, even if an "insulating" layer of gas is formed, there will still be a strong cooling effect whenever liquid nitrogen is involved. The "safe" ...

2

This is a difficult question for many reasons. One reason is likely because most of the introductory thermodynamics textbook problems that we are familiar with from childhood do not involve gravity. To illustrate this difficulty with gravity consider, for example, this snippet from an article in the New York Times Review of Books by physicist/mathematician ...

2

It's an example of adiabatic expansion. If you have a container full of gas and you expand the container, the gas cools. Entropy is preserved. Adiabatic processes preserve entropy. Any decrease in entropy due to lowered energy, and correspondingly fewer possible velocities for the particles, is offset by an increase in entropy due to the expanding volume, ...

1

Let me show you that there is no contradiction by pointing out e.g. that for ordinary expansion periods (that is away from first order phase transitions, decouplings...) the total entropy is actually constant in time while the universe is getting bigger and cooler. Or, going back in time, the universe is getting hotter while S is kept constant. How is this ...

1

There are several ways to approach this problem. If we can estimate the power density achieved in $W/m^2$, then the temperature that can be reached follows from the Stefan-Boltzmann law. First method: 1) Take the total power collected, and see the size it got focused down to. You state the area of the mirror array is 0.6 m$^2$ (roughly), and with power ...

1

Intuitively, For a gas,if you apply heat to the container of gas the kinetic energies of the molecules or atoms increase,means heat added is used in increasing the kinetic energies of the molecules. As we know ,temperature of a gas depends on how fast the molecules of gas moving or vibrating ,so on heating temperature of the gas increases. Now these ...

1

What do you mean by temperature? Let's say you have your gas hooked up to a very tiny thermometer. It has a pointer that jiggles and bumps around. What you'd find is that the pointer fluctuations are mostly related to the mechanical evolution of the thermometer itself, not the system under consideration. But that's beside the point. Temperature is not just ...

1

The Shockley diode equation doesn't distinguish between carrier ($T_{eh}$) and lattice temperature $T$; it assumes that they are in equilibrium, $T_{eh} = T$. Just a word of caution. We can't really say we have a single hot-carrier because temperature is a property of a large number of particles. You can say that you have a hot electron gas which has a ...

1

It doesn't. They'd have to exchange heat through the piston.

1

The best vacuum we can make is about 10^-12 Pa. Atmospheric pressure is about 10^5 Pa, so it's 17 orders of magnitude lower pressure. The best vacuum recorded is the intergalactic void, at about 10^-17 Pa. Even if you managed to remove all matter, there would still be energy from any light or electric fields.

1

The unit $^\circ\mathrm{C}^2$ does make sense. It represents a difference in temperature squared. You are worried that it is invalid because the origin is shifted. This is not a problem though, because we use $\mathrm{m}^2$ all the time, and there is no natural origin at all for positions.

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