Thermal Expansion of a hemisphere

AI Thread Summary
The discussion revolves around calculating the change in interior space of a geodesic dome made of aluminum due to thermal expansion, transitioning from a winter temperature of -18°C to a summer temperature of 30°C. The original volume of the hemisphere was incorrectly calculated, leading to confusion in determining the volume increase. The correct volume formula for a hemisphere was clarified, emphasizing the need to use 2/3 πr^3 instead of halving the volume of a full sphere. The participant acknowledged a significant error in calculating the radius, which affected their final volume calculations. Ultimately, the discussion highlights the importance of accurate formulas and careful calculations in thermal expansion problems.
Samonasuke
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Homework Statement



A geodesic dome constructed with an aluminum framework is a nearly perfect hemisphere; its diameter measures 55.0 on a winter day at a temperature of -18 C.

How much more interior space does the dome have in the summer, when the temperature is 30?


Homework Equations



Linear Expansion: delta L=alpha(L_original)(delta T)
Volume Expansion: delta V=(3)alpha(V_original)(delta T)

The Attempt at a Solution



Ok so I for this question I treated it as a sphere and halved my answer because I though it made sense, not sure if it really mattered

First attempt:
I calculated the original volume of the sphere, which is 47713 m^3. And then I used the volume expansion to calculate the increase in the volume of the sphere which was
delta V=3(2.4*10^-5)(48)(47713) = 164.9L and divide this by 2 and you get 82.4L
But this is wrong

Second attempt:
I tried linear expansion, this way makes near to no sense so I'll abreviate it, I used the linear expansion formula to calculate directly the increase in the radius, and then from there calculated the increase in volume, didnt work

Third attempt: I thought I had it this time
Firsty I calculated the surface area of the sphere = 6361.73
Then calculated the area increase which is delta A=(48)(6361.73)(2)(2.4*10^-5) = 14.66m^2878.53
The calculated the new area =6376.39
Then the new radius, sqrt(6376.39/4Pi) = 22.526 So the new volume of the sphere= 47878.53. Original volume= 47712.93...47878.53-47712.93 =165.6 for a sphere
Volume increase for hemispere= 82.8 m^3 Is this right? Because I'm using masteringphysics and I said 82.5 by mistake and it didnt say I was even close to the answer, I only have one attempt to get this right so I don't want to get it wrong.

Thanks very much for the help, sorry if its a bit convoluted, my first time posting.
 
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Hi Samonasuke! Welcome to PF! :smile:

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Samonasuke said:
A geodesic dome constructed with an aluminum framework is a nearly perfect hemisphere; its diameter measures 55.0 on a winter day at a temperature of -18 C.

I calculated the original volume of the sphere, which is 47713 m^3.

No, the volume will be 2/3 πr3 = 1/12 πd3
 
I was working with a full sphere and then divided my answer in half at the end, anyway I found out where I was going wrong! I can't believe it, I thuoght tha 55/2 22.5 :O Terribly stupid mistake!
 
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