Pendulum with a mass on a container (thermodynamics problem)

In summary, the conversation discusses the conversion of kinetic energy of a pendulum into heat in an adiabatic system. The ideal gas law is used to find the number of moles of gas and a rough estimate of the change in temperature. A calculation error is identified and corrected. The conversation also touches on the variation of entropy and the distribution of energy among different degrees of freedom in the system.
  • #1
ValeForce46
40
3
Homework Statement
A pendulum, with a block of iron (##m=1 kg##, specific heat ##c=448\frac{J}{Kg\cdot K}##) hanging by a thread (mass negligible), is inside a container of volume with rigid and adiabatic walls. Inside the container there's air (to consider as a biatomic ideal gas) at the atmospheric pressure and at the temperature ##T_0=300 K##. You observe the pendulum has a speed of ##v=3.5 \frac{m}{s}## when the block is on the vertical. After some time, the pendulum ceases to swing. Determine:
a) The final temperature of the system at the end of the oscillation.
b) The variation of entropy of the universe at the end of the trasformation.
Relevant Equations
##Q=m*c*(T_f-T_0)##
The kinetic energy of the pendulum ##K=\frac{1}{2}\cdot m\cdot v^2## will turn into heat (entirely).
So both the air and the block of iron will change their temperature.
To find ##n## (moles of the gas) I can use the ideal gas law:
##n=\frac{pV}{RT}=0.9 mol##
Do I have the following equation?
##\frac{1}{2}mv^2=m*c*(T_f-T_0)+n*c_v*(T_f-T_0)## and ##T_f## is my unknown.
But I get ##T_f=300K## (nearly). Where's my mistake?
 
Physics news on Phys.org
  • #2
To get a rough idea of the change in temperature: How many Joules of KE does the block of iron initially have?

Note that ciron = 448 J/(kg K), which tells you that it takes 448 J of energy to raise the temperature of the iron block by just 1 K.
 
  • Like
Likes ValeForce46
  • #3
TSny said:
To get a rough idea of the change in temperature: How many Joules of KE does the block of iron initially have?
##K=6.13 J ##.
So you're silently (not that much :oldbiggrin:) saying that my result ##T_f=300.013 K## is right?
I even doubted my equation.
 
  • #4
ValeForce46 said:
##K=6.13 J ##.
So you're silently (not that much :oldbiggrin:) saying that my result ##T_f=300.013 K## is right?
I even doubted my equation.
Yes, I think your answer is correct. I wanted you to see how you can tell that a very small temperature increase is what you would expect.
 
  • Like
Likes ValeForce46
  • #5
Yeah, you made me realize that. Thanks a lot.
However, for completeness, the variation of entropy of the universe is the variation of entropy of the system, because the walls of the container are adiabatic so there's no variation of entropy of the environment:
##\Delta S_{Universe}=\Delta S_{System}=mc(T_f-T_0)+nc_v(T_f-T_0)=6.1 \frac{J}{K}##
 
  • #6
ValeForce46 said:
##\Delta S_{Universe}=\Delta S_{System}=mc(T_f-T_0)+nc_v(T_f-T_0)=6.1 \frac{J}{K}##
Check this calculation. Note that what you calculated here is ##\Delta E_{int}## of the block and gas. It's not surprising that you got 6.1, since the initial KE of the block is 6.1 J.
 
  • Informative
Likes ValeForce46
  • #7
oops... I did the integral wrong:doh:
##ΔS_{Universe}=ΔS_{System}=m*c*\ln (\frac{T_f}{T_0})+n*c_v*\ln (\frac{T_f}{T_0})=0.02 \frac{J}{K}##
Hope this is right ...
 
  • #8
ValeForce46 said:
oops... I did the integral wrong:doh:
##ΔS_{Universe}=ΔS_{System}=m*c*\ln (\frac{T_f}{T_0})+n*c_v*\ln (\frac{T_f}{T_0})=0.02 \frac{J}{K}##
Hope this is right ...
Looks good.
 
  • Like
Likes ValeForce46
  • #9
Just a side note: For this problem, you can check that about 96% of the initial KE of the iron block ends up in the iron block and only about 4% goes to the air. That kind of surprised me at first. But it makes sense if you think about the equipartition of energy. In this problem, there are a lot more atoms of iron in the system compared to molecules of air. Moreover, for T around 300 K, each atom of iron has effectively 6 "degrees of freedom" for storing internal energy (Law of Dulong and Petit) compared to 5 degrees of freedom for each air molecule. The energy spreads equally among all of the degrees of freedom in the system.
 
  • Informative
Likes ValeForce46
  • #10
ValeForce46 said:
The kinetic energy of the pendulum ##K=\frac{1}{2}\cdot m\cdot v^2## will turn into heat (entirely).
This is an incorrect interpretation. The kinetic energy of the pendulum will not turn into heat. There is no heat transferred to this adiabatic system. The correct interpretation starts with the full presentation of the first law of thermodynamics: $$\Delta U+\Delta (KE)+\Delta (PE)=Q-W=0$$Also there is not change in potential energy of the system. So $$\Delta U=-\Delta (KE)$$
This tells us the correct interpretation, namely, that the kinetic energy of the pendulum will convert directly to internal energy (via viscous dissipation by the air).
 
  • Like
Likes ValeForce46 and TSny
  • #11
Since the temperature hardly changes, the change in entropy is nearly exactly the initial kinetic energy (6.125 J) divided by the temperature, roughly (300 K), or 6.125/300 = 0.0204 J/K.
 

Related to Pendulum with a mass on a container (thermodynamics problem)

1. What is a pendulum with a mass on a container in thermodynamics?

A pendulum with a mass on a container in thermodynamics is a physical system that consists of a container filled with a gas or liquid and a pendulum attached to the top of the container. The pendulum swings back and forth due to the motion of the gas or liquid inside the container, creating a thermodynamic system.

2. How does a pendulum with a mass on a container demonstrate thermodynamics?

A pendulum with a mass on a container demonstrates thermodynamics by showing the transfer of energy between the system (pendulum and container) and its surroundings. The motion of the pendulum is a result of the gas or liquid inside the container expanding and contracting, which is a thermodynamic process.

3. What factors affect the motion of the pendulum in a thermodynamic system?

The motion of the pendulum in a thermodynamic system is affected by several factors, including the mass and length of the pendulum, the type of gas or liquid inside the container, and the temperature and pressure of the system. These factors can impact the frequency and amplitude of the pendulum's swings.

4. How does the temperature and pressure of a thermodynamic system affect the motion of the pendulum?

The temperature and pressure of a thermodynamic system can affect the motion of the pendulum by changing the properties of the gas or liquid inside the container. As the temperature and pressure increase, the molecules in the gas or liquid will move faster and exert more force on the pendulum, causing it to swing with more frequency and amplitude.

5. Can a pendulum with a mass on a container be used to measure thermodynamic properties?

Yes, a pendulum with a mass on a container can be used to measure thermodynamic properties such as temperature and pressure. By measuring the frequency and amplitude of the pendulum's swings, one can calculate the values of these properties using thermodynamic equations. This technique is known as the "pendulum method" and is commonly used in thermodynamics experiments.

Similar threads

  • Introductory Physics Homework Help
Replies
4
Views
814
  • Introductory Physics Homework Help
Replies
17
Views
2K
  • Introductory Physics Homework Help
Replies
12
Views
899
  • Introductory Physics Homework Help
Replies
1
Views
732
  • Introductory Physics Homework Help
Replies
10
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
880
  • Introductory Physics Homework Help
Replies
6
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
298
Back
Top