Find the final temperature of the ball

AI Thread Summary
To find the final temperature of a copper ball with a radius of 1.5 cm that has expanded due to heating, the coefficient of linear expansion for copper is utilized, which is 17 x 10^-6. The coefficient of volume expansion is determined to be three times this value. The volume change is calculated to be approximately 0.54. Initially, there was confusion regarding the application of the equation delta V = BV(delta T), but the correct approach involves adding the room temperature of 22°C to the calculated temperature change. The final temperature of the ball is confirmed to be 410.27°C.
dphoos
Messages
6
Reaction score
0
A copper ball with a radius of 1.5 cm is heated until its diameter has increased by 0.20 mm. Assuming a room temperature of 22°C, find the final temperature of the ball.
 
Physics news on Phys.org
What have you tried?
 
well, I know that the coefficient of linear expansion for copper is 17 x 10^-6, so it's coefficient of volume expansion is 3 times that.. and I know that the volume has changed by about .54. I'm not really sure what to do with this though
 
I assume I'm using the equation delta V = BV(delta T), but when I do this I get a huge number
 
nevermind, I got it. I was subtracting 22 instead of adding it. The answer is 410.27 right?
 

Thread 'Errors and significant figures'

I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
View full post »
Thread 'A cylinder connected to a hanging mass'

Thread 'A cylinder connected to a hanging mass'

Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
View full post »

Similar threads

Change in temperature of air in a closed system
Replies
16
Views
2K
Effect of temperature on surface tension
Replies
2
Views
1K
Question About Momentum: Ball Rolling up a Curved Ramp
Replies
12
Views
1K
Calorimetry - finding the final temperature of a system of ice and water
Replies
7
Views
2K
How Does Newton's Law of Cooling Affect Heat Loss Calculations?
Replies
2
Views
2K
Heat Balance Homework: Find Final Water Temp
Replies
18
Views
2K
Impact of container on final temperature of system
Replies
8
Views
2K
Find the location and length of the final image
Replies
2
Views
836
Mixing two ideal gases with different V, T at constant pressure
Replies
16
Views
3K
What happens to a ball placed on a moving conveyor belt?
Replies
8
Views
3K

Hot Threads

Recent Insights

Back
Top