How to apply the First Law of Thermodynamics to this problem?

In summary, the mass starts at rest and is then stretched and released. It returns to its equilibrium position.
  • #1
vcsharp2003
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Homework Statement


A spring (k = 500 N/m) supports a 400 g mass which is immersed in 900 g of water. The specific heat of the mass is 450 J/kg and of water is 4184 J/kg. The spring is now stretched 15 cm and, after thermal equilibrium is reached, the mass is released so it vibrates up and down. By how much has the temperature of the water changed when the vibration has stopped?

I am ending up with ΔT ( change in temperature of thermodynamic system) and ΔUint (change in internal energy of thermodynamic system) as unknowns and one equation only, so it seems it's not possible to solve for temperature change ΔT.

Homework Equations


Thermodynamics_First_Law.png

heat = mass x specific heat X temperature change

The Attempt at a Solution


[/B]
Let's consider the spring + water as a closed system. Also, let's look at the system starting from when the spring is in water and mass hanging from it in an equilibrium position to when mass returns to it's original position and vibration of spring has stopped. Then, for this system the following points will be true.
  • there could be some heat flow across the system's boundary $$\Delta Q_{in} \neq 0$$
  • the system does not do any work on it's surrounds nor is work done on the system $$W_{by} = 0$$
  • the internal energy of the system should increase as it's temperature will rise $$\Delta U_{int} > 0$$
  • the change in KE of the system is zero since it has no translational motion $$\Delta KE = 0 $$
  • the change in PE of the system is change in spring PE + change in gravitational PE ( change in gravitational PE is zero but change in spring PE is not zero since the spring is less stretched at end of interval being considered i.e. spring PE has decreased)
So, the first law of Thermodynamics becomes
$$(0 - 0) + (0 - \frac{1}{2} \times k \times x^{2}) + \Delta U_{int} = Q_{in} - 0$$
$$(0 - 0) + (0 - \frac{1}{2} \times 500 \times 0.15^{2}) + \Delta U_{int} = Q_{in} - 0$$

Also, by using the second formula under Relevant Equations using specific heat we get another equation. If ΔT = temperature change in K over the time interval being considered, then
$$Q_{in} = 0.9 \times 4184 \times \Delta T + 0.4 \times 450 \Delta T$$

If we combine above two equations then we get the following equation.
$$(0 - 0) + (0 - \frac{1}{2} \times 500 \times 0.15^{2}) + \Delta U_{int} = 0.9 \times 4184 \times \Delta T + 0.4 \times 450 \Delta T - 0$$.

It's impossible to solve for ΔT since we have two unknowns in above equation. I am not sure if I am applying first law of Thermodynamics correctly.

NOTE: However, if Qin is assumed to be zero since heat flows are happening within the system due to fluid resistance against vibration, and ΔUint = 0.9 x 4184 x ΔT + 0.4 x 450 ΔT then by substituting into first law of thermodynamics equation, we get only one unknown of ΔT which we can solve. But, I am not sure if this second approach is correct.
 
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  • #2
The solution comes out in a few lines using simple sums .

How much energy did you start with and where does it go ?
 
  • #3
Nidum said:
The solution comes out in a few lines using simple sums .

How much energy did you start with and where does it go ?
This problem is given at end of the chapter of First Law of Thermodynamics, so I was applying that law. Can you solve this using first law of Thermodynamics?

The energy into the system is 0.9×4184×ΔT+0.4×450 x ΔT. I can equate this to spring PE
.5×500×0.1522 and solve this problem easily.

But, I was looking for a more rigorous way of doing this using first law of thermodynamics, if it's possible.
 
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  • #4
You were correct that Q is equal to zero, for the reason that you gave. You did a pretty good job of analyzing the rest of the problem. (I'm not so sure that the change in gravitational potential energy is zero in this process, but, apparently, they expected you to make this assumption.)
 
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  • #5
The mass starts at rest in the water and is then stretched and released. It returns to its equilibrium position.
Only the vibrational KE is dissipated in the water, the temperature of the water and the mass rise.
 
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  • #6
Chestermiller said:
You were correct that Q is equal to zero, for the reason that you gave. You did a pretty good job of analyzing the rest of the problem. (I'm not so sure that the change in gravitational potential energy is zero in this process, but, apparently, they expected you to make this assumption.)

I am thinking that the initial equilibrium position of mass should be same as final equilibrium position of mass according to free body diagram below, which will mean initial and final gravitation PE are equal.
When mass is in initial rest position ( not when its stretched by 15 cm but before that when it's hanging by itself from spring in water) or when it has finally stopped after vibration in water, then the following equation is satisfied.

Tension in spring = weight + upthrust due to displaced water
kx = weight + upthrust due to displaced water

Since k ( spring constant), weight of mass and upthrust due to displaced water are going to be same in both these positions, so x (amount by which spring is stretched) is going to be same in both positions. Therefore, we can say that gravitation PE are same in both these positions.

Also, then equation from first law of Thermodynamics should be different as below.

$$\Delta KE = 0 \text {, } \Delta PE = 0 \text { for both gravitational and spring PE, } Q_{int} = 0 \\ \text {but } W_{by} = -\frac{1}{2}kx^{2} \text { since spring was initially stretched by 15 cm and so work was done on the system}$$The first law of Thermodynamics would then become as below.
$$(0) + (0) + \Delta U_{int} = 0 - (-\frac{1}{2}kx^{2})$$

We can substitute ΔUint = 0.9×4184×ΔT+0.4×450ΔT in above equation.

Mass_in_Water_attached_to_spring.png
 
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  • #7
vcsharp2003 said:
I am thinking that the initial equilibrium position of mass should be same as final equilibrium position of mass according to free body diagram below, which will mean initial and final gravitation PE are equal.
When mass is in initial rest position ( not when its stretched by 15 cm but before that when it's hanging by itself from spring in water) or when it has finally stopped after vibration in water, then the following equation is satisfied.

Tension in spring = weight + upthrust due to displaced water
kx = weight + upthrust due to displaced water

Since k ( spring constant), weight of mass and upthrust due to displaced water are going to be same in both these positions, so x (amount by which spring is stretched) is going to be same in both positions. Therefore, we can say that gravitation PE are same in both these positions.

Mass_in_Water_attached_to_spring.png
If you are going to take as the initial state the equilibrium rest location of the mass, then you need to include the work done in lowering the mass as part of the process, and this will also involve gravitational (buoyancy) effects. If you take as the initial state, the lower mass location with the stretched spring, then there is some mass of water that is at a higher elevation than at equilibrium and the mass itself which is at a lower elevation than at equilibrium.
 
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  • #8
Chestermiller said:
If you are going to take as the initial state the equilibrium rest location of the mass, then you need to include the work done in lowering the mass as part of the process, and this will also involve gravitational (buoyancy) effects. If you take as the initial state, the lower mass location with the stretched spring, then there is some mass of water that is at a higher elevation than at equilibrium and the mass itself which is at a lower elevation than at equilibrium.
Yes, I realized that after posting my last message and have modified the post accordingly. Thanks for your great help. I think the upthrust is being ignored in this problem else the volume of the mass would be mentioned..
 
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  • #9
vcsharp2003 said:
Yes, I realized that after posting my last message and have modified the post accordingly. Thanks for your great help. I think the upthrust is being ignored in this problem else the volume of the mass would be mentioned..
Yes. Very perceptive.
 
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  • #10
Chestermiller said:
You were correct that Q is equal to zero, for the reason that you gave. You did a pretty good job of analyzing the rest of the problem. (I'm not so sure that the change in gravitational potential energy is zero in this process, but, apparently, they expected you to make this assumption.)
I came back to this problem and was confused about one concept. In this problem the sum of heat flow into the water and heat flow into the mass is taken as change in internal energy of the system. However, the sum of heat flows into water and mass should be factored into ##\Delta Q## and not into the change of internal energy ##\Delta U## unless ##\Delta U= \Delta Q## for the mass and water system
 
  • #11
vcsharp2003 said:
This problem is given at end of the chapter of First Law of Thermodynamics, so I was applying that law. Can you solve this using first law of Thermodynamics?

The energy into the system is 0.9×4184×ΔT+0.4×450 x ΔT. I can equate this to spring PE
.5×500×0.1522 and solve this problem easily.

But, I was looking for a more rigorous way of doing this using first law of thermodynamics, if it's possible.
What @Nidum proposed is still a thermodynamic 1st law solution.
 
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  • #12
vcsharp2003 said:
Qin is assumed to be zero since heat flows are happening within the system due to fluid resistance against vibration
That can be considered as the Qin - the change of state of the spring from stressed state to relaxed state, with the process releasing heat to the system.
 
  • #13
vcsharp2003 said:
Qin is assumed to be zero since heat flows are happening within the system due to fluid resistance against vibration
That can be considered as the Qin - the change of state of the spring from stressed state to relaxed state, with the process releasing heat to the system.
 
  • #14
256bits said:
What @Nidum proposed is still a thermodynamic 1st law solution.
I have the impression that Thermodynamics First Law is ##\Delta U = \Delta Q - \Delta W##. Any calculations involving specific heat capacity must be directly used as ##\Delta Q## and not as ##\Delta U##.

Also, the mass attached to spring gains in gravitational potential since the equilibrium final position is higher than stretched position of the mass energy while the spring potential energy is decreased since spring is less extended when equilibrium position is reached.
 
  • #15
256bits said:
That can be considered as the Qin - the change of state of the spring from stressed state to relaxed state, with the process releasing heat to the system.
I think this is just a simple application of conservation of energy rather than First Law of Thermodynamics. Trying to fit the First Law of Thermodynamics into this problem is an overkill and unnecessary. It's simply that the spring mass system has a net negative work, which means that this work is lost as heat energy, and that's all there is to this problem. Positive work on a system means gain of energy and negative work means a work is converted to another form of energy.
 
  • #16
vcsharp2003 said:
I think this is just a simple application of conservation of energy rather than First Law of Thermodynamics. Trying to fit the First Law of Thermodynamics into this problem is an overkill and unnecessary. It's simply that the spring mass system has a net negative work, which means that this net amount of work in magnitude is lost as heat energy, and that's all there is to this problem.
What is your definition of "heat energy?"
 
  • #17
Chestermiller said:
What is your definition of "heat energy?"
It's energy in transit.
 
  • #18
vcsharp2003 said:
It's energy in transit.
But the problem statement suggests that there is no thermal energy interchange between the surroundings and the system in this problem.
 
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  • #19
I think they should have stated what is the ##\Delta T ## of the water after a very long time. The dissipated stored spring energy is traveling to into parallel sinks with different thermal resistivities. I would suspect the spring would stop and the change in temperature for each mass would be initially different. After a "long time" thermal equilibrium would be reached between the water and the mass which would be the ##\Delta T ## the question was after?
 
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  • #20
Chestermiller said:
But the problem statement suggests that there is no thermal energy interchange between the surroundings and the system in this problem.
Ok, that part I get.

I seem to be stuck with a certain concept which may be why I'm not understanding this clearly. According to my understanding ##Q= m \times c \times \Delta T##, gives us the heat flow and not the change in internal energy ##\Delta U_{int}##. If I applied the idea that heat flows are zero into/out of the mass-spring + water system then ##\Delta U_{int}## would still be not known. But here, we use the specific heat formula to determine ##\Delta U_{int}##, when we know the formula gives us the heat flow ##Q## and not ##\Delta U_{int}##.

If you can explain to me why ##Q= m \times c \times \Delta T## can be used to get ##\Delta U_{int}## for mass-spring + water system, then I will understand it clearly.
 
  • #21
vcsharp2003 said:
Ok, that part I get.

I seem to be stuck with a certain concept which may be why I'm not understanding this clearly. According to my understanding ##Q= m \times c \times \Delta T##, gives us the heat flow and not the change in internal energy ##\Delta U_{int}##. If I applied the idea that heat flows are zero into/out of the mass-spring + water system then ##\Delta U_{int}## would still be not known. But here, we use the specific heat formula to determine ##\Delta U_{int}##, when we know the formula gives us the heat flow ##Q## and not ##\Delta U_{int}##.

If you can explain to me why ##Q= m \times c \times \Delta T## can be used to get ##\Delta U_{int}## for mass-spring + water system, then I will understand it clearly.
That equation is used to describe the change in internal energy given an amount of heat ##Q## being absorbed or released by the body. It’s just a simplification of the first law, ## Q = \Delta U_{int} = m c_p \Delta T##

The ##Q## in this case is the amount of work initially done to compress the spring. It’s released into the system via friction and is absorbed by the mass (block + water) in the system.

I think you might be hung up on what we commonly encounter in early physics/chemistry with the calorimetry problem. This is in effect this problem with the chemical energy released in reaction replaced by a stretched spring with energy released via non-conservative forces.

In the first law ##Q - W = \Delta E## the heat can go to a number of energy sinks on the RHS in the system. ##Q = m c_p \Delta T## is a very special case of the first law.
 
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  • #22
vcsharp2003 said:
Ok, that part I get.

I seem to be stuck with a certain concept which may be why I'm not understanding this clearly. According to my understanding ##Q= m \times c \times \Delta T##, gives us the heat flow and not the change in internal energy ##\Delta U_{int}##. If I applied the idea that heat flows are zero into/out of the mass-spring + water system then ##\Delta U_{int}## would still be not known. But here, we use the specific heat formula to determine ##\Delta U_{int}##, when we know the formula gives us the heat flow ##Q## and not ##\Delta U_{int}##.

If you can explain to me why ##Q= m \times c \times \Delta T## can be used to get ##\Delta U_{int}## for mass-spring + water system, then I will understand it clearly
That's because, in thermodynamics, $$\left(\frac{\partial U}{\partial T}\right)_V=mC_v$$ not ##Q= m \times c \times \Delta T##.
 
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  • #23
Chestermiller said:
That's because, in thermodynamics, $$\left(\frac{\partial U}{\partial T}\right)_V=mC_v$$ not ##Q= m \times c \times \Delta T##.
So, heat flow for a constant volume system always equates to change of internal energy of the system . Is that what the differential equation is meaning?
 
  • #24
vcsharp2003 said:
So, heat flow for a constant volume system always equates to change of internal energy of the system . Is that what the differential equation is meaning?
No, the differential equation is saying much more that that. We are saying that, in thermodynamics, the definition of heat capacity has been surreptitiously changed from the earlier definition in terms of heat to the new definition in terms of internal energy or enthalpy (without informing you of the change).
 
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  • #25
Chestermiller said:
No, the differential equation is saying much more that that. We are saying that, in thermodynamics, the definition of heat capacity has been surreptitiously changed from the earlier definition in terms of heat to the new definition in terms of internal energy or enthalpy (without informing you of the change).
All this is confusing to me.

The theory given in the textbooks is completely different and easy to comprehend.

The following makes more sense to me, when looking at constant volume process, which seems similar to the equation you gave.

"At constant volume, dV = 0, δQ = dU (isochoric process).

A system undergoing a process at constant volume implies that no expansion work is done, so the heat supplied contributes only to the change in internal energy.

The heat capacity obtained this way is denoted ## C_{V}##. The value of ## C_{V}## is always less than the value of ## C_{P}##.
##C_V< C_P##".

https://en.wikipedia.org/wiki/Heat_...t volume, dV = 0, δQ = dU (,),-Calculating CP
 
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  • #26
vcsharp2003 said:
All this is confusing to me.

The theory given in the textbooks is completely different and easy to comprehend.
Sorry about that. I can't help it if they changed the definition of heat capacity and forgot to tell the students about it. This has been a source of confusion to students for generations.
 
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  • #27
vcsharp2003 said:
All this is confusing to me.

The theory given in the textbooks is completely different and easy to comprehend.

The following makes more sense to me, when looking at constant volume process, which seems similar to the equation you gave.

"At constant volume, dV = 0, δQ = dU (isochoric process).

A system undergoing a process at constant volume implies that no expansion work is done, so the heat supplied contributes only to the change in internal energy.

The heat capacity obtained this way is denoted ## C_{V}##. The value of ## C_{V}## is always less than the value of ## C_{P}##.
##C_V< C_P##".

https://en.wikipedia.org/wiki/Heat_capacity#:~:text=displaystyle C_{P}.}-,At constant volume, dV = 0, δQ = dU (,),-Calculating CP
According to Smith and Van Ness, Introduction to Chemical Engineering Thermodynamics, Chapter 2:
"There are in fact two heat capacities in common use for homogeneous fluids; although their names belie the fact, both are state functions, defined unambiguously in relation to other state functions:

Heat capacity at constant volume$$C_V\equiv\left(\frac{\partial U}{\partial T}\right)_V\tag{2.20}$$Heat Capacity at constant pressure $$C_P\equiv\left(\frac{\partial H}{\partial T}\right)_P\tag{2.21}$$
.... Consider now the case in which the volume varies during the process, but is the same at the end as at the beginning. Such a process can't rightly be called one of constant volume, even though ##V_2=V_1## and ##\Delta V=0##. However, changes in state functions or properties are independent of path and are, therefore, the same for all processes which lead from the same initial to the same final conditions. Hence, property changes for this case may be calc[ulated from the equations for a truly constant-volume process leading from the same initial to the same final conditions. For such processes, ... ##\Delta U=\int{C_vdT}##, because U, Cv, and T are all state functions or properties. On the other hand, Q does depend on path, and the equation ##Q=n\int{C_VdT}## is valid for Q only for a constant-volume process."
 
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  • #28
First I think the question to be answered will be clearer if instead of
vcsharp2003 said:
By how much has the temperature of the water changed when the vibration has stopped?
it became "By how much has the temperature of the water changed when the vibration has stopped and the system has reached a new equilibrium temperature?"
Second, I think it is safe to consider this as a simple calorimetry problem in which the relevant masses are isolated from the environment and all energy transformations are internal.
Third, the internal energy of the system initially is in the form of thermal energy of the water and 400-g mass and the elastic potential energy ##U_{el}=\frac{1}{2}kx^2## in the spring (assumed massless).
Fourth, after the spring is released, the elastic potential energy is divided unequally between the water and the mass. A simple equation involving the given specific heats describes this mathematically and can easily be solved to find ##\Delta T## because it is the same for both water and 400-g mass.

Additional assumption: The simple harmonic motion is underdamped and the mass eventually ends up at its equilibrium position.

N.B. In post #15 OP suggests that this problem involves conservation of (total) energy and that using the first law of thermodynamics is "overkill." OP might wish to consider that conservation of total energy, conservation of mechanical energy, the work-energy theorem and the first law of thermodynamics are all manifestations of the same idea: "The net change is what goes in less what comes out; if nothing goes in or comes out, the net change is zero." It works with bank accounts too.
 
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  • #29
erobz said:
In the first law Q−W=ΔE the heat can go to a number of energy sinks on the RHS in the system. Q=mcpΔT is a very special case of the first law.
I see. What will be W according to the situation you have explained?

If W = 0 then I can understand why in this problem, ##\Delta U_{int} = mc\Delta T##. But, why would W = 0 beats me.
 
  • #30
vcsharp2003 said:
I see. What will be W according to the situation you have explained?

If W = 0 then I can understand why in this problem, ##\Delta U_{int} = mc\Delta T##. But, why would W = 0 beats me.
I'm not sure, but my gut instinct is I think you have to be careful about selecting the system boundary.

If you make a boundary only around the water (excluding the mass) we can write the first law for it. It absorbs some heat, gains some bulk kinetic energy, and it does work (through friction force ##f_r##) on the mass\sping to bring it to rest. Then you can isolate the mass\spring and write the first law from its perspective. It too absorbs some heat, and does work on the water through friction force ##f_r##. A friction force is doing work on both of them that in that in the end ( after thermal equilibrium is once again reached between the water and mass\spring) is converted to waste heat. I think in the end all three of these things combine to give you:

$$ \int f_r ~ds = \frac{1}{2}k x^2 = Q_{{abs}_w} + Q_{{abs}_b} = \left( m_w c_{p_w} + m_b c_{p_b} \right) \Delta T $$

Thats the best idea I can come up with.
 
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  • #31
erobz said:
Thats the best idea I can come up with.
I think trying to force yourself to use the First Law of Thermodynamics for such a problem is unnecessary. It can simply be solved by using the law of conservation of energy that we learn in the chapter of Work, Power and Energy. The spring mass system is losing some spring potential energy and that energy lost is converted to thermal energy. That's all there is to this problem.
 
  • #32
vcsharp2003 said:
But, why would W = 0 beats me.
The system of spring, 400-g mass and water is isolated. No external work or heat crosses the system's boundary. So W = 0. If you think the work is not zero, what exactly is it that does work on the system?
 
  • #33
kuruman said:
If you think the work is not zero, what exactly is it that does work on the system?
That seems correct. There is no external work happening since the work done by the water on the mass is not external work, but like internal work. Right? I'm not sure if the work done against water by the mass is to be included in thermodynamic work W.
 
  • #34
vcsharp2003 said:
That seems correct. There is no external work happening since the work done by the water on the mass is not external work, but like internal work. Right?
Right. One part of the system may do mechanical work on another part, but the other part does equal and opposite work on the first part. This is guaranteed by Newton's 3rd law.
 
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  • #35
kuruman said:
Right. One part of the system may do mechanical work on another part, but the other part does equal and opposite work on the first part. This is guaranteed by Newton's 3rd law.
That's a beautiful explanation. I think I am getting how to apply thermodynamics to this situation.
 
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