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

[ISU20CpreE]

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.

 

[HallsofIvy]

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?

 

[ISU20CpreE]

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]

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]

I get 2.65 cm^3 as an answer.

is this answer corect?