Why is U used instead of Q to define temperature?

[iScience]

in class we learned the definition of temperature to be $\frac{1}{T}$=$\frac{∂S}{∂U}$

i don't understand why it's U as opposed to Q. afterall, Q is the only form of energy that contributes to temperature isn't it? If i take a bathtub of water and i swirl my arm in it, i just gave it some work, and the temperature will have gone up, but that temperature increase is due to the work being converted to Q right? So why is it the entropy per unit U as opposed to the entropy per unit heat (this would be a phonon right?)

 

[dauto]

in class we learned the definition of temperature to be $\frac{1}{T}$=$\frac{∂S}{∂U}$

i don't understand why it's U as opposed to Q. afterall, Q is the only form of energy that contributes to temperature isn't it?

No, work also contribute

If i take a bathtub of water and i swirl my arm in it, i just gave it some work, and the temperature will have gone up, but that temperature increase is due to the work being converted to Q right?

No, not right. The work is being converted into internal energy U

So why is it the entropy per unit U as opposed to the entropy per unit heat (this would be a phonon right?)

I don't understand what connection is there with a phonon.

 

[iScience]

I don't understand what connection is there with a phonon.

well, intermolecular vibrations of say, a solid (i don't know about the other phases, I'm guessing this only applies for solids), are quantized and i was taught therefore that the unit of heat (vibration) could be said to be quantized as phonons.

 

[stevendaryl]

in class we learned the definition of temperature to be $\frac{1}{T}$=$\frac{∂S}{∂U}$

i don't understand why it's U as opposed to Q. afterall, Q is the only form of energy that contributes to temperature isn't it? If i take a bathtub of water and i swirl my arm in it, i just gave it some work, and the temperature will have gone up, but that temperature increase is due to the work being converted to Q right? So why is it the entropy per unit U as opposed to the entropy per unit heat (this would be a phonon right?)

I think this is in the wrong location. This question is classical physics, not quantum physics.

But $T$ is not a function of $Q$ alone. When you compress a gas, it gets hotter, but not because you've transmitted any heat to it.

 

[dauto]

well, intermolecular vibrations of say, a solid (i don't know about the other phases, I'm guessing this only applies for solids), are quantized and i was taught therefore that the unit of heat (vibration) could be said to be quantized as phonons.

You're confusing heat Q with internal energy U. The vibration of molecules (whether quantized or classical) is part of the internal energy U which is a function of state. The heat Q is not a function of state and has nothing to do with internal vibrations. Heat Q is energy transferred between two objects (systems) due to their difference in temperature. When energy moves from an object with a high temperature to an object at lower temperature by either radiation, or conduction, or convection, that energy is considered heat. The energy inside of an object related to its thermal state is called internal energy and is represented by the letter U.

 

[iScience]

I think this is in the wrong location. This question is classical physics, not quantum physics.

:O sorry! i thought i clicked classical not quantum.. no wonder i couldn't find it there.. could a moderator reading this perhaps move this thread to the classical section?

You're confusing heat Q with internal energy U. The vibration of molecules (whether quantized or classical) is part of the internal energy U which is a function of state. The heat Q is not a function of state and has nothing to do with internal vibrations. Heat Q is energy transferred between two objects (systems) due to their difference in temperature. When energy moves from an object with a high temperature to an object at lower temperature by either radiation, or conduction, or convection, that energy is considered heat. The energy inside of an object related to its thermal state is called internal energy and is represented by the letter U.

So then.. heat is defined as the TRANSFER of energy? i don't understand, if heat is inherently defined as the 'transfer' of something, well.. all transfers (as far as i know) have a rate associated with them, so why isn't it d(something)/dt?

it would make sense that the 'somethig' is energy, such that it's an energy transfer rate;
but i know this is incorrect because Q has units of energy and not power.

So then, is Q inherently an integral of the transfer rate of energy w/ respect to dt? ie..

Q=$\int$Pdt

the units do match up...

When energy moves from an object with a high temperature to an object at lower temperature by either radiation, or conduction, or convection, that energy is considered heat.

say i have an object that does not interact with gamma rays, ie it doesn't absorb it. Now say that i shine gamma rays through this object of really high intensity say... I=10,00000GW/m². if you consider an imaginary box around this object, would it then be appropriate to say that this object has a lot of heat? this just seems counter intuitive to me; again, if heat is the 'transfer' of energy, and radiation counts as "heat", then this object should then have a high internal energy even though it may be sitting at.. 4kelvin, right?

 

[dauto]

:O sorry! i thought i clicked classical not quantum.. no wonder i couldn't find it there.. could a moderator reading this perhaps move this thread to the classical section?




So then.. heat is defined as the TRANSFER of energy? i don't understand, if heat is inherently defined as the 'transfer' of something, well.. all transfers (as far as i know) have a rate associated with them, so why isn't it d(something)/dt?

it would make sense that the 'somethig' is energy, such that it's an energy transfer rate;
but i know this is incorrect because Q has units of energy and not power.

So then, is Q inherently an integral of the transfer rate of energy w/ respect to dt? ie..

Q=$\int$Pdt

the units do match up...

Correct. Heat is the energy transferred over a period of time, not the transfer rate which would indeed have units of power.

say i have an object that does not interact with gamma rays, ie it doesn't absorb it. Now say that i shine gamma rays through this object of really high intensity say... I=10,00000GW/m². if you consider an imaginary box around this object, would it then be appropriate to say that this object has a lot of heat? this just seems counter intuitive to me; again, if heat is the 'transfer' of energy, and radiation counts as "heat", then this object should then have a high internal energy even though it may be sitting at.. 4kelvin, right?

If the object doesn't absorb the radiation than no energy is being transferred.The radiation goes right through the object. There is no heat. Right now as we speak (so to speak) there are billions of neutrinos coming from the sun's core and passing right through your body (and mine). These neutrinos are not absorbed and do not heat your body.

 

[iScience]

Correct. Heat is the energy transferred over a period of time, not the transfer rate which would indeed have units of power.

If the object doesn't absorb the radiation than no energy is being transferred.The radiation goes right through the object. There is no heat. Right now as we speak (so to speak) there are billions of neutrinos coming from the sun's core and passing right through your body (and mine). These neutrinos are not absorbed and do not heat your body.

i got it now! thank you so much!