Volume Expansion: 15.5cm Sphere Heated from 40°C to 210°C

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The discussion focuses on calculating the volume change of a 15.5 cm diameter quartz sphere when heated from 40°C to 210°C. The initial volume calculated is approximately 15598.53 cm³, and the temperature change, Δ(T), is confirmed to be 170°C. Participants discuss the coefficient of thermal expansion (β) for quartz, which is noted as 0.000001. One user reports calculating a volume change of 2.65 cm³ and questions its correctness. The conversation emphasizes the importance of correctly applying the thermal expansion formula and using accurate values for β.
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A quartz sphere is 15.5 cm in diameter. What will be its change in
volume if it is heated from 40°C to 210°C?
Ok so what I did was use the diameter of the sphere and use the volume formula of the sphere and I got 15598.53 cm^3. Then what I did was use this formula \Delta(V) = \beta V_o\Delta(T) knowing this formula I plug in the informationg given in the problem, but the part I don't understand is if its \Delta(T) shouldn't it be the final tamperature minus the initial? Giving me 170 C. I can't do that part I need help.
 
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Yes, \Delta(T) is the change in temperature, 170 degrees Centigrade. What do you mean you "can't do that part"? Have you looked up \beta for quartz?
 
HallsofIvy said:
Yes, \Delta(T) is the change in temperature, 170 degrees Centigrade. What do you mean you "can't do that part"? Have you looked up \beta for quartz?
YEs i did but i really can't get the answer the Beta for quartz is 0.000001 i don't get the answer right. What can i do?
 
ISU20CpreE said:
YEs i did but i really can't get the answer the Beta for quartz is 0.000001 i don't get the answer right. What can i do?

I get 2.65 cm^3 as an answer.
 
ISU20CpreE said:
I get 2.65 cm^3 as an answer.

is this answer corect?
 
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