What Is the Final Temperature of a Heated Copper Ball?

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SUMMARY

The final temperature of a heated copper ball can be accurately determined using the volume expansion formula rather than linear expansion. The discussion highlights that for a copper ball with a radius of 1.4 cm and a diameter increase of 0.22 mm, the correct approach involves the equation (delta)V = 3*alpha*V*(delta)T. It is crucial to convert measurements to meters and use the radius for calculations to find the change in volume (delta)V accurately.

PREREQUISITES
  • Understanding of thermal expansion concepts, specifically volume expansion.
  • Familiarity with the coefficient of linear expansion for materials, particularly copper.
  • Basic knowledge of geometry, specifically the relationship between radius and diameter.
  • Ability to perform unit conversions, particularly from centimeters to meters.
NEXT STEPS
  • Study the principles of thermal expansion in solids, focusing on volume expansion equations.
  • Research the coefficient of linear expansion for various materials, including copper.
  • Practice converting measurements between different units, especially in thermal physics contexts.
  • Explore real-world applications of thermal expansion in engineering and material science.
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Students in physics or engineering, educators teaching thermal dynamics, and professionals involved in material science or mechanical engineering will benefit from this discussion.

FishieKissie06
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A copper ball with a radius of 1.4cm is heated until its diameter has increased by 0.22mm. Assuming a room temperature of 22*C, find the final temperature of the ball.

well what i did was i used linear expansion for this and i didnt get it right. i used (delta)L=(alpha)initial L*(delta)Temperature...solving for temperature
 
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Well in order to do this... you must use volume expansion since it is not a rod gaining length... so you must use the equation... (delta)V = 3*alpha*V*(delta)T and for (delta)V you can substitute in V1-V2
and remember that you must have it in raidus and not diameter when finding (delta)V so... since the radius then increases by (d/2) you have to remember to convert to meters for both of the radius...
 

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