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Andre

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Is this a first grade question and what is the correct answer? 50 degrees Celsius? I think not. Somehow, I believe the answer to be 53,8 degrees Celsius. Am I right?

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- #1

Andre

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Is this a first grade question and what is the correct answer? 50 degrees Celsius? I think not. Somehow, I believe the answer to be 53,8 degrees Celsius. Am I right?

- #2

G01

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- #3

Andre

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- #4

G01

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Use this equation to find out how much heat both start out with. C is the specific heat of the object. Then find an average of the heats and work out what each metal' temperature would be at those heats. I think that should work.

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G01

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sry i wanted uercase delta there, woops

- #6

Andre

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Again this is no homework question. Just some brain exercise. Is this correct?

- #7

Doc Al

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Where did you get this formula from? Common sense and symmetry should tell you that the final temperature will be right in the middle between the two initial temperatures.

To figure it out, realize that the net heat flow must be zero, so:

[tex]m_1 c_1 \Delta t_1 + m_2 c_2 \Delta t_2 = 0[/tex]

But since the two objects are identical:

[tex]m c \Delta t_1 + m c \Delta t_2 = 0[/tex]

[tex]\Delta t_1 = -\Delta t_2[/tex]

Since the objects reach thermal equilibrium, they have the same final temperature. Which, per the above, must be right in the middle.

To figure it out, realize that the net heat flow must be zero, so:

[tex]m_1 c_1 \Delta t_1 + m_2 c_2 \Delta t_2 = 0[/tex]

But since the two objects are identical:

[tex]m c \Delta t_1 + m c \Delta t_2 = 0[/tex]

[tex]\Delta t_1 = -\Delta t_2[/tex]

Since the objects reach thermal equilibrium, they have the same final temperature. Which, per the above, must be right in the middle.

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- #8

Andre

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Which, per the above, must be right in the middle.

But in the middle of what? The middle of the temperature or the middle of the energy?

It's excruciating obvious that 50C is the middle of the temperature but perhaps that that 53.8 degrees Celsius is the average (middle) of the energy. no? After all it's the energy that we have a conservation law of, it's not about about conservation of temperature.

We do not measure the temperature but we measure the length of a fluid column that depends on the volume of that fluid in a bulb at the underside of the thermometer. We define the temperature to be proportional to the volume of that fluid. What is temperature again? Wasn't it the same as average speed of the individual molecules.

Now what I'm not sure of, but assume that the volume of that fluid is directly proportional to the average speed of the molecules. Everything is linear up to now.

But the energy the heat is proportional with the square of the speed of the molecules. So double temperature is four times more energy No?

I'm quite puzzled here. Did we overlook something so obvious in the most plain basics of physics or am I way of track?

- #9

Doc Al

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Both!Andre said:But in the middle of what? The middle of the temperature or the middle of the energy?

No. The average (translational) KE of the molecules is proportional to the temperature, not the temperature squared.But the energy the heat is proportional with the square of the speed of the molecules. So double temperature is four times more energy No?

The amount of energy transferred as heat is proportional to the change in temperature.

- #10

Andre

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But the ticks on the thermometer are on a lineair scale.

- #11

Doc Al

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Sounds good to me.Andre said:If the KE would be proportional to the temperature and you add a fixed amount of energy for each degree C to the bulb of the thermometer, the speed of the molecules of the fluid would increase with the square root of the delta energy.

Why would you assume that? You could make a handwaving argument that the average amplitude of the molecular vibrations in a solid is proportional to theWe assume that the speed of those molecules is proportional to the expansion of the volume of the fluid (it's probably more complex), hence the column of the thermometer would grow with the square root of the added energy.

Right. And they are used to measureBut the ticks on the thermometer are on a lineair scale.

- #12

Andre

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You could make a handwaving argument that the average amplitude of the molecular vibrations in a solid is proportional to the energy

Good one. But this is true for harmonic oscilations. Just thinking. I can imagine that molecar motion is anti harmonic. Molecules speeding up when nearing others due to interacting gravity. So the speed could be highest on the end of the trajectory before the collision. If energy and temperature and thermal expansion are truly proportional then I could also image that the decrease in density is a factor for the square root relation to the average speed.

- #13

Bystander

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V ~ V_{0}(1 + 3εΔT) ,

where V

The capillary cross-sectional area in liquid in glass thermometers is a function of ε

Any other questions?

- #14

Andre

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[tex]pV=nRT[/tex]

or

[tex]\frac{p_1V}{T_1}=\frac{p_2V}{T_2}[/tex]

Now if T2 is double T1 and the volume V remains constant in a fixed container, then the pressure P2 is double P1 as well.

Pressure is force per area. So double pressure is double force as the area is constant.

the force is the sum of the momentums of the individual molecules (mv) per second:

[tex]F=ma=\sum \frac{m_iv}{s}[/tex]

So doubling the force means doubling the individual speeds of the molecules as m and time remains constant, which brings me to square one. Energy is proportional to the square of the speed. So, am I forced to conclude that doubling the temperature means four times more energy? What am I doing wrong.

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russ_watters

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- #16

russ_watters

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No, temperature is defined to be proportional to kinetic energy, which is why it makes for a linear scale with volume of a fluid. There is no discrepancy there.Andre said:We define the temperature to be proportional to the volume of that fluid. What is temperature again? Wasn't it the same as average speed of the individual molecules.

What you are imagining just isn't reality: those quantities - energy, temperature, and thermal expansion, really are proportional.But this is true for harmonic oscilations. Just thinking.I can imaginethat molecar motion is anti harmonic. Molecules speeding up when nearing others due to interacting gravity. So the speed could be highest on the end of the trajectory before the collision. If energy and temperature and thermal expansion are truly proportional then I could also image that the decrease in density is a factor for the square root relation to the average speed.[emphasis added]

- #17

Doc Al

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So far, so good.Andre said:Now if T2 is double T1 and the volume V remains constant in a fixed container, then the pressure P2 is double P1 as well.

Pressure is force per area. So double pressure is double force as the area is constant.

The force exerted on the walls is the averagethe force is the sum of the momentums of the individual molecules (mv) per second:

[tex]F=ma=\sum \frac{m_iv}{s}[/tex]

What you are doing wrong is assuming that the rate that particles hit the wall remains the same. But, since the particles are moving faster, they hit the walls more often. The change in momentum of a single collision is proportional to speed, but the number of collisions per second is also proportional to speed. Thus the pressure is proportional to theSo doubling the force means doubling the individual speeds of the molecules as m and time remains constant, which brings me to square one. Energy is proportional to the square of the speed. So, am I forced to conclude that doubling the temperature means four times more energy? What am I doing wrong.

So it all makes sense: Double the temperature and the pressure doubles, since both are proportional to speed

- #18

coburg

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- #19

Andre

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That's it indeed. More collisions. Thanks all, I was lost for a moment.

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