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13 votes
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Does color temperature limit how much a laser of a given wavelength can heat a target?

The argument for heating by the Sun is that if the target object is hotter than the Sun it would heat the Sun rather than the Sun heating it. We could try to apply this argument to a laser, but how ...
John Rennie's user avatar
6 votes

Does color temperature limit how much a laser of a given wavelength can heat a target?

Neither the Sun nor a laser are objects in thermodynamic equilibrium The second law of thermodynamics, which prohibits the flow of heat from a colder to a warmer object, applies to objects that have ...
Roger V.'s user avatar
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6 votes

Does color temperature limit how much a laser of a given wavelength can heat a target?

Assuming that the target is a black body. The real critical factor affecting heat transfer is emit energy per unit area described by Stefan–Boltzmann law $M=\sigma T^4$. The temperature increase ...
et al.'s user avatar
  • 151
4 votes

Why is Sun's energy entropy low on Earth?

It is said that we can utilize the energy from the sun on earth because it's low entropy Actually, we can utilise energy from the Sun because the temperature of the Sun is higher than the temperature ...
gandalf61's user avatar
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4 votes

Does color temperature limit how much a laser of a given wavelength can heat a target?

No, your reasoning does not apply to a laser. You can use Wien’s law to relate the peak emission wavelength to the temperature of a black body emitter (a perfect absorber/emitter across all ...
FTT's user avatar
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4 votes

Does color temperature limit how much a laser of a given wavelength can heat a target?

No, it is totally unrelated. Look at the National Ignition Facility (at LLNL). It has a laser, I do not know what its frequency is (probably a small multiple of $c/1064\,$nm, but it heats a pellet to ...
JEB's user avatar
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3 votes
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Enthalpy at constant pressure/volume

Enthalpy is defined as $H=U+pV$ irrespective of what process the system is subjected to. It, enthalpy $H$, is a state function and in equilibrium thermostatics its definition has nothing to do with ...
hyportnex's user avatar
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3 votes
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Why won't there be any transfer of heat energy when ice at 0°C is in contact with water at 0°C in a closed container?

The other answers have hammered quite well that There is no such thing as heat inside a thing; heat is the name for the spontaneous transfer of energy due to temperature differences and so if it is ...
naturallyInconsistent's user avatar
3 votes

Why won't there be any transfer of heat energy when ice at 0°C is in contact with water at 0°C in a closed container?

As others have said, systems don't contain heat but contain internal energy. So, let's pretend your textbook used the correct term. To socratically answer your question -- why there is not heat ...
jwimberley's user avatar
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2 votes
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Heating an object with black body radiation to above the temperature of the source

Thought experiment 1 A is a large black body at temperature $T_A$ in empty space. B is a small black body some distance away at the lower temperature $T_B$. I expect that B will gradually be warmed ...
Roger V.'s user avatar
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2 votes

Heating an object with black body radiation to above the temperature of the source

While a lot of the other answers touch on the various aspects of the problem, the real reason this can't be done is more subtle. First, it's not true that you can't use sunlight to heat something to ...
A Nejati's user avatar
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2 votes

Why won't there be any transfer of heat energy when ice at 0°C is in contact with water at 0°C in a closed container?

"If there is no transfer of heat between the two bodies placed in contact, they are said to be at the same temperature, but it doesn't mean that they have equal amount of heat in them." ...
Bob D's user avatar
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2 votes

Why won't there be any transfer of heat energy when ice at 0°C is in contact with water at 0°C in a closed container?

Heat transfer only occurs when there is a temperature difference.The heat flux 'q' s.t by$$\vec q =-\lambda \nabla T$$where $T$ is temperature and '$\lambda$' is coefficient, and if $T$ is same in ...
UnnamedUser's user avatar
1 vote
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Contradiction in the Carnot Cycle?

You are talking about two different temperature differences. The first is the temperature difference between the heat engine and the thermal reservoirs during the isothermal expansion and compression ...
Bob D's user avatar
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1 vote

Contradiction in the Carnot Cycle?

There is no contradiction. The Carnot cycle has 4 stages consisting of two isothermal stages, one at a temperature $T_1$ and the other at a temperature $T_0 < T_1$, say, and two adiabatic stages. ...
hyportnex's user avatar
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1 vote

Besides the 2nd law of thermodynamics, what laws of optics prevent the temperature of the focal point of lens from being hotter than the light source?

My easiest argument was that the 2nd law of thermodynamics prevents this from happening because heat can't flow passively from a place of lower energy/entropy to a place of higher energy/entropy. He ...
Ján Lalinský's user avatar
1 vote

When is the temperature relevant for a quantum many body system?

If the system is at $T=0$, all particles should be at the ground state and no excited state is possible. However, this is clearly not true for an arbitrary quantum system, given that if we solve ...
Roger V.'s user avatar
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1 vote

The usage of temperature in quantum mechanics

Many-body quantum mechanics is also often referred to as Quantum statistical physics, which is a more telling name in the sense that it points out that we are considering systems where one has to take ...
Roger V.'s user avatar
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1 vote

Heating an object with black body radiation to above the temperature of the source

To add to the existing answers, we can quantitatively show that $T_B$ will never exceed $T_A$ in your "Thought experiment 2" using Planck's law. The key idea is that the filter must be ...
Roger Yang's user avatar
1 vote

Heating an object with black body radiation to above the temperature of the source

The other answers came close but does not seem to have hammered this into you: This attenuates the longer wavelengths so that the transmitted radiation has the same spectrum as a hotter object. This ...
naturallyInconsistent's user avatar
1 vote

When is the temperature relevant for a quantum many body system?

I will write my answer by addressing some of your statements in your question, and elaborate a bit around them. You start by saying: The Hamiltonian of the system is typically of the form (expression ...
Marius Ladegård Meyer's user avatar
1 vote

When is the temperature relevant for a quantum many body system?

The question in the title, the "when" question, is basically impossible to answer. I mean, it is only when your level of precision required for whatever experiment it is you are doing, ...
naturallyInconsistent's user avatar
1 vote

Heating an object with black body radiation to above the temperature of the source

I expect that B will gradually be warmed but not to above the temperature of A. Probably not even close to it. While B is receiving some radiation from A, B will be radiating to empty space in all ...
BowlOfRed's user avatar
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1 vote

Heating an object with black body radiation to above the temperature of the source

Filtering longer wavelengths will not make the spectrum equivalent to one of a hotter object. You'll simply be cutting out the region where the colder object peaks and leaving the tails that have a ...
agaminon's user avatar
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1 vote
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Why does capillary in gas thermometer have a temperature and pressure gradient?

Arguably, Zemansky isn't explicating any important thermodynamic point here but instead listing possible sources of error when using a certain device associated with thermodynamics—in other words, he'...
Chemomechanics's user avatar

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