# Tag Info

3

Good question. The rate of temperature increase scales as the power absorbed by the food divided by mass of the food. So to understand your question, you need to understand how power is absorbed. There is a finite amount of power in the microwaves being produced. These microwaves bounce around in the metal cage where you put your food, until they come ...

0

Just write it out like this: $t = \frac{e}{\frac{m}{shc}}= \frac{J}{\frac{kg}{\frac{J}{kg\cdot{}^\circ{}C}}} = \frac{J}{J \cdot kg \cdot \frac{1}{kg\cdot^{\circ}C}} = \frac{1}{\frac{1}{^{\circ}C}} = ^{\circ}{\rm{}C}$

3

When multiplying or dividing units, all you need to do is put the units in the numerator or denominator (wherever they appeared) of the answer. So: $$[e/M]={J\over kg}$$ $$[M/shc]={kg\over{J\over kg^oC}}={kg^2\,^{\circ}\rm C\over J}$$ But this is not the correct way of analyzing your units. You have $t = e / M / shc = e / (M * shc)$ The units of this are: ...

1

Integrability of the inexact differential $\delta Q$ is a law of nature. Although in general Pfaffian differential forms like $\delta Q$ are not integrable, second law of thermodynamics guarantees that an integrating factor always exists and it is $1/T$ in all cases, $T$ being the absolute temperature.

1

Your method seems correct. Here are the details: $Q_{Heat Water}=m\times c\times (T_2-T_1)=100 \times 1 \times 66 = 6600$ calories From 1 gram of steam, $$Q=L_v+c\times(100-T_2)=540+1 \times 10=550$$ Therefore, grams of steam needed $=\frac{6600}{550}=12$ grams

1

The best way to see this is to realize that the zero heat capacity is a quantum effect. Classically, the heat capacity does not go to zero. Quantally what happens is that at low enough temperatures all the particles are in their lowest possible energy states. To get even one particle into a higher energy state requires a small but finite energy ...

0

The air flow, the block's efficiency, etc. do have an effect. However, it would be considered "a transient" effect. After you allow "enough time," only the temperature difference, is what matters.

1

Let's assume there is excellent thermal conductivity from the heater to the block, and from the block to the inner surface of the radiator. This is a convective radiator. The rate of transfer of heat energy depends linearly on the difference in temperature between the block and the air (since you're holding other things fixed). The whole system will reach ...

1

1 - yes the zero temperature limit is not reachable, so you can't measure the heat capacity at zero temperature, what this calculation tell you is that if you measure at smaller and smaller temperatures you will see that C converges towards zero 2- No the reversibility of the path is not important as the entropy is an exact differential

0

Simply because the rate of flow of heat slightly increased when you added more water: Heat is transferred very quickly to the kettle (which I assume is made out of metal), as metal is a good conductor of heat. Air is a bad conductor, so no heat enters the water through the air. Water is a worse conductor of heat than metal (it also has a pesky habit of ...

5

Double the ammount of water does not need doulbe the ammount of time to heat, since while the energy needed is doubled indeed, losses due to vaporization and radiation from the kettle should be approximately constant. You can plot the time needed for a given ammount of water to boil and try to fit a function into that. With two data points you can manage to ...

0

Resistance is simply (resistivity X length)/area Since your resistivity is material dependent and length is also fixed, you can manipulate the area. Decreasing the cross section area of the wire does mean that you are effectively increasing the resistance. You have to optimize the parameters so you get the max out of it. And as the first answer points ...

3

Suppose we have a source of electrical energy, say a battery, that puts out 100 Volts. It is connected through wires with a total resistance of 1 ohm to a heater with a resistance of 99 ohms. The battery sees a total resistance of 100 ohms, and thus pushes 1 Ampere of current through the circuit. The battery is delivering energy at 100 Watts The Power ...

3

To get the local heating you will need some measure of the optical density of the material and an estimate of the local intensity of the light. You can probably look-up an estimate of the optical density, or if you need more precision measure it yourself. Starting from a known intensity, ray optics will give you an easy estimate of the position dependent ...

22

Yes, it's 11% hotter today than yesterday. Of the three temperature scales you discussed, only the Kelvin scale allows meaningful ratios to be calculated. Dividing two temperatures expressed in Celsius or Fahrenheit is simply a mistake. There are numerous physical examples where it makes sense to multiply or divide by a Kelvin temp, e.g., the ideal gas law ...

0

One way is to find out the internal change energy of the system and infer the heat transfer to the system from that and the work done: $$\delta Q_\text{to}=dU-\delta W_\text{on}.$$ If you have a handle on the system's entropy, on the other hand, then you can use the Gibbs relation, $$\delta Q_\text{to}=TdS,$$to find the heat delivered. In general, though, ...

1

Assuming the surface of the metal remains smooth, the reflection from it will be specular and the metal will look shiny regardless of the temperature. However the amount of light metals absorb, instead of reflecting, generally increases with increasing temperature because you get more scattering of the conduction electrons by lattice vibrations. So the metal ...

0

If you have a path on $p-V$ diagram that is parametrized by some parameter $x$, so that $p=p(x),\,V=V(x)$, then: $$dU=\delta Q+p(x)dV(x)$$ here $Q$ is the total heat received by the system (it is negative if system releases heat). I write $\delta Q$ to indicate that $Q$ is not a function of state, and $\delta Q$ is not a full differential. Assume now that ...

0

AB-isothermal. $\Delta W=Q$ ; area under the curve depicts the work done , ie. heat intake. $\Delta W=n\Bbb RT\ln\dfrac{V_2}{V_1}$ BC-isobaric. Can't be calculate directly from curve. otherwise use ,$Q=\Delta W + \Delta U$ where $\Delta W=P(\Delta V)$ and $\Delta U=\dfrac f2 p\Delta V$ CD-isochoric. $Q=\Delta U=\dfrac f2 V\Delta p$ . DA-adiabatic . ...

0

It depends where the aluminum foil is situated. For example I use aluminum foil behind a wood stove to reflect the heat to the room instead of heating the wall. So, does it mean that a more shiny aluminium foil will reflect more light and thus make the room more cooler as compared to less shiny foil? You can see from Crazy Buddy's answer that the more ...

1

First, there's no perfect reflector nor absorber. In fact - even Aluminium does absorb some radiation (by which it gets heated, can be noticed at incident high frequency radiation). One more thing is that aluminium foils are designed in a way to reflect light. Here's the Wiki article quote... Aluminium foil has a shiny side and a matte side. The shiny ...

0

As you've asked about part (a) now, and shown a little work in your comment, I'll just answer regarding that part. Initially, there is a nonzero pressure inside the syringe. Presumably, this is balanced by an equal and opposite force from the surrounding air. Now, that part of the force won't change after placing the mass on the syringe, so you still have ...

0

The input stream does not only have a thermal energy - it also has a mechanical one. Mechanical energy can be used for work, and the gas temperature is easily changed by work - in adiabatic processes it rises when gas is pressed, and falls when gas is able to expand. This gives a general idea why this tube could work and at the same time not be a Maxwell's ...

-1

The short answer is they don't cool nearly as well as ice The harvard food and science team did an experiment for this exact scenario. The results were: As the ice melts it drops the temperature of Whiskey to nearly -4C, where as the whiskey stones barely got the drink close to zero. The reason has already been explained nicely in Manishearth's answer.

9

TL;DR: Whiskey stones work by absorbing heat from the whiskey in an attempt to reach thermal equilibrium1. As Thomas mentioned, ice has three cooling effects: Ice itself takes 2.11 kilojoules of heat per g to have its temperature increased by 1 degree (Celsius). This number is known as "specific heat capacity" Ice takes 334 kilojoules of heat per kg at 0 ...

5

Whiskey stones aren't necessarily designed to keep the drink cold, instead they are designed to allow flavor profiles to come out in the drink that might not be present at room temperature. Some whiskeys open up at a slightly cooler temperature and using stones allows you to experiment without diluting the flavor of the beverage. There are better math ...

48

Ice cubes have three distinct cooling effects: The cube, initially at sub-zero temperature, absorbs some heat to reach fusion point (0⁰C). The cube absorbs more heat to switch phase: it takes some energy to turn 1 kg of ice at 0⁰C into 1 kg of liquid water at 0⁰C. The water absorbs some heat to become warmer than 0⁰C. The three effects occur more or less ...

11

A fundamental principle of thermodynamics is that heat flows from warm places to cold ones, through either convection, conduction or radiation, and it will continue to do so until the temperature equalizes across the system. The stones are colder than the whiskey when you put them in the glass, so as the system heads towards equilibrium, the whiskey gets ...

0

Absolutely. This is how a microwave ovens work (at least when you put metal in them). Microwave ovens emit electromagnetic radiation in the microwave region (roughly $2.4 \, \mathrm{GHz}$) Metal plates in a microwave oven act as antennas and the electromagnetic radiation induces a huge electric current. If you are looking for specific details on how ...

0

Are you sure that the first block of code is correct? V2 should have been multiplied by V. You get correct values here only because you assumed V=1. No, the difference in volumes isn't equal (or even in linear proportion) to difference in densities because they have an inverse relation. In terms of densities you have: \$\alpha = {m (\frac 1 \rho - \frac 1 ...

1

Your own (David Cary's) answer is good, but there is one thing you've overlooked, which might make a big difference in some situations. This is simply that there might periods of the day in which you're not in the building, and ideally you'd like to turn the air conditioning off during those times. The advantage of a low thermal mass in this case should be ...

1

That depends on what you mean by "heat". If you mean energy then you can calculate the energy required to cause the temperature change using the following equation: Energy = mass * specific heat capacity * temperature change (Q = mcθ).

1

As for the question of whether anything can be hotter than the sun. The Sun is composed of plasma, an energetic phase of matter in which electrons get ripped off of atoms, and electrons and ions coexist in something that might best be described as an ionized gas. According to this wiki page, the so-called Z machine has achieved temperatures on the order of ...

0

It depends on many factors such as the reentry velocity of the object, its shape (cone-spherical, etc.), what the planet's atmosphere is made of, whether it enters at some shallow angle and also the altitude where there's density variations in atmosphere, etc. Googling on this, could return you a lot of results. And, all results matched a certain value. ...

0

Let's see some issues on "what would happen if the answer was Yes???" and especially for a highly massive object like a bubble... Our Sun would exert more gravitational pull on objects than if it were a white dwarf. (Going deeper) The core of stars would exert more gravity on the outer layers leading to extreme compression of celestial objects. In fact, ...

1

A good example would be heating a tin can of water using a Bunsen burner. Initially the flame produces radiation which heats the tin can. The tin can then transfers heat to the water through conduction. The hot water then rises to the top, in the convection process. The atmosphere would be another example. The atmosphere is heated by radiation from the ...

0

The artificial rocks would be just as radioactive. By melting the waste into the rock, you only add some of the rock's matter to the shielding. So you invested energy to create a shape that's handy, but equally dangerous. This is what's actually done. The spent fuel is molten into glass which is then cast into a concrete shell (see Wiki: ...

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