Question on the effects of size of bulb on thermomters

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Increasing the size of the liquid-in-glass bulb affects thermometer performance by increasing the time taken to register temperature changes due to greater thermal mass and resistance. The scale markings must also be adjusted to maintain accuracy, as the volume change alters the relationship between temperature and liquid expansion. To keep the scale identical while increasing bulb size, the bore diameter should be proportionally increased. If the bulb radius is doubled, the volume increases eightfold, necessitating a bore radius increase by the cube root of eight to maintain scale consistency. Overall, these changes highlight the importance of bulb size on thermometer functionality and accuracy.
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1. Hi, I've got this question that's asking for the effects of increasing the size of the liquid -in-glass bulb. I think they're asking for at least 3 points. 3. L think its supposed to accentuate the increase in volume of the liquid, and it increases the time taken to measure the temperature. Are these correct?Thanks for all your help!
 
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Would you conserve or change the graduations on the thermometer, and if so, why?
 
I think the graduation and scale should remain the same.
 
Why do you think so?
 
wait... if they remain the same, then the scale would be wrong. so, the scale should change, to maintain the correctness of the thermometer.
 
So... Any idea what the effects would be?
 
I agree that if we change the bulb volume this should affect the time for temperature changes to be registered. That's effect 1.

You agreed the need to change the scale markings to deal with the increased volume of liquid. That's effect 2.

By the way, any idea how we could increase the size of the bulb but keep the scale markings identical?

There are obvious changes such as a change in the mass, centre of gravity, and inertial moment of the thermometer, but these do not seem particularly relevant.
 
I think if you wanted to keep the scale, what you could do would be to increase the diameter of the bore. Therefore, the diameter of the bore should be in direct proportion with the increase in the bulb, to keep the scale.
 
Therefore, the diameter of the bore should be in direct proportion with the increase in the bulb

Let's be clear about this. If I double the radius of the bulb, do I double the radius of the bore?
 
  • #10
I'm would think so...
makes sense i suppose.
 
  • #11
If I double the radius of the bulb, by how much does the volume of the bulb increase?

If I double the bore radius of the tube, by how much do I increase the internal volume for a given length of tube? (We want to keep the temperature graduations the same distance apart).
 
  • #12
What's point three?
 
  • #13
What's point three?

Would you like to suggest one? I can think of loads but I'm not sure the OP wants them.
Increased manufacturing cost would be one example!
 
  • #14
Carid said:
Would you like to suggest one? I can think of loads but I'm not sure the OP wants them.
Increased manufacturing cost would be one example!

The keyword is "effect"--"what's the effect?"

I was thinking of three characteristics of a larger volume: dV/dT, thermal mass, and thermal resistance. The last two have the same effect.
 
  • #15
Phrak said:
The keyword is "effect"--"what's the effect?"

I was thinking of three characteristics of a larger volume: dV/dT, thermal mass, and thermal resistance. The last two have the same effect.

Just wondering, What's that same effect. Longer time to measure to measure temperature?
 
  • #16
MforceXXII said:
Just wondering, What's that same effect. Longer time to measure to measure temperature?

Yes, the amount of time. Greater thermal mass; greater time. Greater thermal resistance; greater time.
 
  • #17
i agree that the volume of the bore should be proportional to the volume of the bulb to keep the same scale. However if you double the radius of the bulb, the volume wil grow 8fold.
If you want the volume of the bore to grow proportionally it will need a radius 8^(1/2) bigger since the volume of a sphere is r^3*pi*(4/3) and the volume of a cilinder pi*r^2*h
 
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