Matching a graph to an experiment

In summary, the conversation involves a student from France seeking help with physics tasks while their teacher is away. They discuss three experiments involving a small ball, a capacitor, and heated water and try to determine which one is correct. The student also brings up the idea of considering heat loss to the environment, and they discuss the use of Newton's law of cooling and other factors such as convection and radiation. Eventually, they arrive at a solution for the third experiment using the equation ##\dot Q_{\rm lost} =k*(T-20)##, with k estimated to be 5 W/K.
  • #1
franceboy
51
0
hi there,
i am from France, so excuse my mistakes languagewise, please.
We got some long-time tasks in physics because our teacher is away. One of them is really hard. I hope you can help me a little bit.
Match the following graph to one of these experiments.
Since the graph is very small and you can not see it on computer really well, i have wrote the look-up table of the graph.
exp1: The collision of a small ball on a planar table which is located 20 cm over the ground.
y is the height of the ball in cm and x is the time after the collision in ms.
exp2: A capacitor with the capacity of 6.8 μF is charged with a resistance of 3.3 MΩ. At the beginning he has a voltage of 20V. y is the voltage in V and x is the time in s.
exp3: 100 ml water are heated electrically with the power of 350 W. The water has the temperature of 20°C at the beginning. y is the temperature in °C and x the time in s.


Ok now my ideas:
exp1: The ball is small and has a low velocity so that we can neglect the air resistance. Then the formula is h(t)= 0.2m + v*t + 1/2*g*t^2, and v is the start velocity of translation in the direction of y. We can determine v from the table: v≈ 7 m/s. I have calculated some values but after 80 ms they differ a lot from the ones of the table. So i thaught that exp1 is wrong.
exp2: The formula is U(t)= 20V+(U-20V)*(1-e^(-t/(R*C))) with U as maximal voltage. The slope in the start point would be U´(0)= 1/(R*C)*(U-20V)≈ 0.7 V/s. The value 0.7 V/s was determined with the table. But than U would be lower than 40V which is wrong.
exp3: My strategy is to use the energys:
Q= m*c*ΔT= 350W*t. However T(t) is a linear function then which differs from the values.

Consequently every experiment would be false but our teacher said that one of them is wright!

THANK YOU in advance :smile:
 

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  • #2
Your discussion of option 3 leaves out loss of heat to surroundings.

Edit: That said, the shape is clearly a negative exponential, which suggests option 2. I agree the RC value doesn't fit, though. It would need to be about 85s.
 
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  • #3
Bonjour France, and welcome to physicsforums.

Haru points you in the right direction. Make a simple assumption about the heat lost to the environment, such as Q_lost = X(T-20) with X a constant.
 
  • #4
Thanks for your quick answers.
Yeah, i had the idea of considering the loss of heat. I thought I could use Newton`s law of cooling but I did not find a function which includes the cooling while something is heated. Another option i thought of is to use the process of convection, conduction and radiation. However, that would be very complicated...
BvU, you mentioned that Q_lost is equal to k*(T-20), i thought Q_lost would be something like (T-20)*e^(-k*t) because of Newton´s law.
It would be nice if you could explain this.
 
  • #5
franceboy said:
Thanks for your quick answers.
Yeah, i had the idea of considering the loss of heat. I thought I could use Newton`s law of cooling but I did not find a function which includes the cooling while something is heated. Another option i thought of is to use the process of convection, conduction and radiation. However, that would be very complicated...
BvU, you mentioned that Q_lost is equal to k*(T-20), i thought Q_lost would be something like (T-20)*e^(-k*t) because of Newton´s law.
It would be nice if you could explain this.
I think BvU meant the rate of loss of heat.
You are right that in general heat loss is complicated. Newton's law really only applies to forced convection. In natural convection the air flow increases with temperature difference, so the rate of cooling will increase faster than linearly:
- conduction: linear
- convection: some higher power
- radiation: 4th power
In general,
##\dot {\Delta \theta} = P/S - \Delta\theta f(\theta)##
where Δθ is the excess temperature over ambient, P is the applied power, S the specific heat of the object.
Since f is unknown here, how can you check whether the data might fit such an equation?
 
  • #6
I would estimate that f is a konstant k. Then I could detrermine k with a tangent on the graph, so that Iwould have a differential equation which I can solve. Is that right? Otherwise I would need some more help.
 
  • #7
Okay i got 0.005 for k and the function is T(t)=-P/k*e^(-k*t/(m*c)+P/k+20 and the values of this function are really good. However is this the physically right function?
 
  • #8
Sorry for the confusion: yes, ##\dot Q_{\rm lost} =k*(T-20)##.

You shouldn't ask "is that right", but follow your own idea. Much more satisfaction from discovering by yourself that it is a good one !

[edit] posts crossed. I found k = 5 W/K and indeed an almost perfect fit. Somewhat unlikely because one would expect k to increase with temperature (not absolutely sure, but I seem to remember: more movement of the liquid, evaporation, etc.)
 
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1. How do you match a graph to an experiment?

To match a graph to an experiment, you need to first understand the variables involved in the experiment. Then, you can plot the independent variable on the x-axis and the dependent variable on the y-axis. Finally, you can compare the shape and trend of the graph to the expected results of the experiment to determine if it is a good match.

2. What is the purpose of matching a graph to an experiment?

Matching a graph to an experiment allows us to visually represent the data collected during the experiment. It helps us to analyze and interpret the results more easily, and to identify any patterns or trends that may exist.

3. What if the graph does not match the expected results of the experiment?

If the graph does not match the expected results of the experiment, it could indicate that there were errors or inconsistencies in the data collection process. In this case, it is important to review the experimental procedure and identify any potential sources of error.

4. Can a graph be used to make predictions about future experiments?

Yes, a graph can be used to make predictions about future experiments. By analyzing the trends and patterns in the graph, we can make educated guesses about how the variables may behave in similar experiments in the future.

5. Are there any limitations to matching a graph to an experiment?

One limitation of matching a graph to an experiment is that it only provides a visual representation of the data and may not capture all the complexities of the experiment. It is important to also consider other factors and analyze the data in more detail to fully understand the results of the experiment.

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