# Tag Info

18

The two "no" answers you've already received are correct for all practical purposes. In real-world cases there can be a difference though. The difference depends on when the refrigerator decides to cycle on and cool. If the fridge cycles on a timer or based on heat energy then there will be a difference due to the added heat capacity. The outside of the ...

12

The heat loss (power) at a particular temperature is the same. So, No - the cooling needed to maintain the thing cold stays roughly the same. However, the empty fridge has lower total heat capacity. So, it will get warm faster in the absence of power. So, it is worthwhile to fill your fridge and freezer with bottle of water a few days before a big storm ...

10

Diamond dust (or dust of any other material) won't conduct heat anywhere close to as well as the solid material. At a molecular level the dust isn't in very good contact with other grains of dust. There is plenty of separation and air in between the particles that will retard heat conductivity. If you were to compress the dust so significantly that it did ...

9

No. The rate of cooling must simply match the rate of heating, and heating rate depends only on the temperature difference you want to establish and on the thermal conductivity and surface area of the walls. More stuff in the refrigerator would give it a higher heat capacity, so that it wouldn't warm up so much when the door is opened. However, it will take ...

7

It's a steady state. If there were a pressure gradient, there would be net force on the gas (ignoring gravity). There's no net force here because the air isn't accelerating. Thus the pressure is constant. The number density varies across the box inversely to the temperature so the ideal gas law holds.

5

For metals there is a connection between the thermal conductivity and electric conductivity (Wiedemann–Franz law). However specific heat is not directly related. This is because electric and thermal conductivity are due to the electrons, however the specific heat is mostly due to the ion vibrations (phonons). Despite "classical" intuition electrons ...

5

I can't make this a comment since I don't have enough reputation. The metal box itself can absorb some of the heat (by conduction) and then give out energy in the form of electromagnetic radiation. If you want to do some real physics with such a system, you could idealize the box as a black body and continue.

5

Thermal conductivity relates to the propagation of heat, whereas electrical conductivity relates to the effective propagation of electric charge. In the case of thermal conductivity, not only the electrons play a role in the conduction but also phonons or magnons contribute to it. The electrons only play a significant role in heat conductivity in metallic ...

4

In a liquid mixture such as ethanol-water, both components vaporize to some extent. If the combined vapor pressure of the two equals the external pressure, say 1 atm, the mixture will boil. The components DO NOT boil separately. Further, the composition of the vapor and the composition of the liquid will be different from each other. This is the basic ...

4

Heat is the thermal motion of particles. Hot object's atoms vibrate more than cold object's atoms. Heat is transfered by 3 main ways: Conduction: Heat flows from hot objects to cold objects. If you have an electric stove, the heat flows from the coils to the pan. Convection: Heat flows by bulk motion of a fluid. If you heard "hot air rises" this is the ...

4

Your first answer is the correct one. The temporal auto-correlation function (or the time correlation function) of a fluctuating quantity $A(t)$ is $$C_{AA}(\tau) = \langle A(t) A(t+\tau)\rangle$$ This is a measure of how correlated $A(t)$ is to its value at another time $t+\tau$. The Green-Kubo formula that connects a response function (the electrical ...

4

By conductor of heat, do you mean that it is bad at transferring heat via conduction? Or that it is just bad at transferring heat? First, a picture of the molecular structure of an oil: Conduction Oil is a liquid. Heat transfer by conduction requires strong bonds between the molecules, so that a vibration(heat) travels down the line. With liquids, this ...

4

Your first assumption is not really correct. The induction cooker creates an electromagnetic field with a frequency in the range of 10 kHz to 100 kHz. The coil used in the cooker looks as follows: The coil and the frequency are choosen in such a way that the field does not extend more than a few centimeters along the axis of the setup. Now when you put a ...

4

Both bodies of water will heat till 100 C (as well as the vessel). Then, the outside one will start boiling by taking latent heat from the surroundings, Now, we only have heat transfer due to conduction/radiation when there is a temperature difference (or an emf, but that's irrelevant). It's mainly conduction we need to consider, anyways. Convection won't ...

4

Water boils at 100° C and atmospheric pressure. If you try to boil water in a closed container, a small amount of steam will form; this will increase the pressure, and the rest of the water will not boil (unless the container bursts or leaks). This explains why water will not boil inside a plastic sac. I don't completely understand the set-up with the ...

4

Your confusion lies within your perception of natural length in the Young's modulus formula. When we say strain=$\Delta L/L$, the $L$ refers to the natural length of the rod at a given temperature. So, if the rod is not clamped, and we increase the temperature, there is no deviation from natural length at that temperature (as we can define natural length of ...

3

I'm sure it depends on the mug and the temperature of the coffee, but most of the time I bet that evaporative cooling from the top is the dominant source of heat loss. That's just based on experience--like it stays hot much longer when you cover the top, and much shorter when you blow on the top. I also think that it cools at a similar rate in my ceramic mug ...

3

I should see the whole article to give a proper conclusion, but I can tell you this from my experience as an experimental solid state physicist: When you are doing contact measurements and you have unusual properties, e.g. non-ohmic transport or dielectric response, it is possible that these unusual properties are artifact, that is not due to bulk of the ...

3

Materials that become superconductors have different thermal conductivities in the normal and the superconducting state. When in the superconducting state, starting a current larger than the critical one will drive such a material to its normal state, thereby altering its thermal conductivity.

3

Short answer: because food is more dense than air, and thus can retain "coldness" longer, once it has been cooled down. All refrigerators leak temperature, mainly through joints. Problem is, in an empty refrigerator, all you have is cold air (which is the main heat transmitter inside a fridge). Look for heat capacity. Thermodynamics is really an eye opener ...

3

Consider two regions $R_1$ and $R_2$ separated by an interface consisting of a planar surface. Let $\mathbf n_{21}$ denote the unit normal vector along the interface pointing from volume $1$ to volume $2$. If energy (in the form of heat conduction) is being transferred between these two systems, then this transfer has a direction in the sense that heat can ...

3

Depending on the what level of precision you need for your experiment, you may need to distinguish the grade of copper that you are using to conduct the experiment. Pure, oxygen-free copper are usually used for high vacuum environments, and thermoexpansion curves for those can be found - I found one from UCSD at http://aries.ucsd.edu/LIB/PROPS/PANOS/cu.html. ...

3

I am not a physicist either. As I understand it, heat can be lost by conduction, by convection and by radiation, The purpose of the bottle is to reduce all three. If you half the amount of liquid, the question is whether you also half the loss of heat, or do more or less. Analysis is difficult because the weak part of the bottle is the cork. If it is ...

3

I don't have a thermos flask to hand, otherwise I'd do the experiment (the only sure way to answer :-). In the absence of experimental data I'd guess that the half full flask will cool faster. The heat flow will be roughly proportional to the temperature difference between the inside of the flask and the ambient temperature outside. The constant of ...

3

It depends on the volume of your fridge versus the amount of bottles you have. Either way it is always best to keep the bottles that are being cooled as far apart as possible in order to maximize surface area to release heat. If your fridge can barely fit all the bottles at once then it would be better to put a few bottles at a time in the fridge, wait for ...

3

Without doing the analysis, I would think that a cooling system is more effective at extracting heat from a warm container than from a cold one. For a fridge, the effectiveness (or coefficient of performance) is $Eff=Q_c/W$ is the ratio of the heat removed from the cold source (the fridge) to the energy used for the purpose. It increases with the ...

3

The general form of the heat equation would be $$\frac{\partial T}{\partial t} + (\vec{u} \cdot \nabla) T =a \nabla^2T + S$$ The first term is the time derivative, the second term convection, the third term diffusion of heat and the last term is the source term, which can be anything. However, normally I would expect radiation to come into the equation ...

3

Matter above absolute zero will radiate (electromagnetic) energy no matter what. This is due to the motion of atoms (specifically charged subatomic particles) in the energized matter. Conduction between two bodies in thermal contact is only one means of transferring energy - it is different than radiation. The earth does not need to be in contact with ...

2

The heat equation for these kind of problems (assume 1D), reads $$\frac{\partial T}{\partial t} = a \frac{\partial^2 T}{\partial x^2}$$ Where of course $T$ is temperature, $t$ is time, $x$ is position and $a$ the thermal diffusivity: $a=\frac{\lambda}{\rho c_p}$ (respectively thermal conductivity, density and heat capacity. This equation can be derived ...

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