Matching a graph to an experiment

  • Thread starter franceboy
  • Start date
  • #1
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:
 

Attachments

Answers and Replies

  • #2
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
34,338
5,891
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.
 
Last edited:
  • #3
BvU
Science Advisor
Homework Helper
2019 Award
13,539
3,265
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
51
0
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
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
34,338
5,891
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
51
0
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
51
0
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
BvU
Science Advisor
Homework Helper
2019 Award
13,539
3,265
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.)
 
Last edited:

Related Threads on Matching a graph to an experiment

Replies
2
Views
758
Replies
2
Views
1K
  • Last Post
Replies
1
Views
4K
Replies
1
Views
1K
  • Last Post
Replies
9
Views
4K
Replies
1
Views
10K
  • Last Post
Replies
7
Views
1K
  • Last Post
Replies
4
Views
909
Replies
4
Views
463
Top