2 more thermal expansion questions

In summary, the solution to the first question is to find the temperature at which the diameter of the steel rod and brass ring are the same, which can be determined using the linear equation delta L=alpha(Lo)(delta T). For the second question, the period of a pendulum depends on its length, so on a hot day (30 C), the clock would run faster due to the increased length of the pendulum, resulting in a gain of time in a 24 hour period.
  • #1
F|234K
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0
1.) a steel rod is 3 cm in diameter at 25 C. a brass ring has an interior diameter of 2.992 cm at 25 C. at what common temperture will the ring just slide into the rod?

for this Q, i am not too sure which equation to use. i think it's the volume change one, which is delta V= beta(Vo)(delta T)

but i don't know how the temperature and the diameter is going to fit in...

2.) a grandfather's clock is calibrated at a temperature of 20 C. the pendulum is a think brass rod with a heavy mass attached to the end. on a hot day, when the temperature is 30 C, does the clock run fast or slow? how much time does it gain or lose in a 24 hour period?

this Q...i just have no idea how to start...

thanks in advance, any suggestions/comments are greatly appreciated.
 
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  • #2
F|234K said:
1.) a steel rod is 3 cm in diameter at 25 C. a brass ring has an interior diameter of 2.992 cm at 25 C. at what common temperture will the ring just slide into the rod?

for this Q, i am not too sure which equation to use. i think it's the volume change one, which is delta V= beta(Vo)(delta T)

but i don't know how the temperature and the diameter is going to fit in...
Forget about volume; the rod will just slide into the ring when their diameters are the same. You need to find the [itex]\Delta T[/itex] that will have them end up with the same diameter.
2.) a grandfather's clock is calibrated at a temperature of 20 C. the pendulum is a think brass rod with a heavy mass attached to the end. on a hot day, when the temperature is 30 C, does the clock run fast or slow? how much time does it gain or lose in a 24 hour period?
Hint: How does the period of a pendulum depend on its length?
 
  • #3
for number one, are u suggesting taht i should use the linear equation instead of the volume one?

linear equation being delta L=alpha(Lo)(delta T)
 
Last edited:
  • #4
"Hint: How does the period of a pendulum depend on its length?"

taht the thing i don't no...so i can't do the Q...
 
  • #5
F|234K said:
for number one, are u suggesting taht i should use the linear equation instead of the volume one?
Absolutely.

Regarding the pendulum: Look it up! :smile:
 
  • #6
thanks alot.
 

1. What is thermal expansion?

Thermal expansion is the tendency of a material to increase in size when heated and decrease in size when cooled.

2. What causes thermal expansion?

Thermal expansion is caused by the increase in kinetic energy of the particles in a material when heated, causing them to vibrate and move further apart.

3. How does thermal expansion affect materials?

Thermal expansion can cause materials to change in size, shape, and density, which can lead to structural changes and potential damage.

4. Are all materials affected by thermal expansion?

Yes, all materials are affected by thermal expansion to some degree. However, the degree of expansion may vary depending on the type of material and its composition.

5. How can thermal expansion be controlled or minimized?

Thermal expansion can be controlled or minimized by choosing materials with low coefficients of thermal expansion, using proper design and construction techniques, and implementing compensation mechanisms such as expansion joints or thermal breaks.

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