Question

There are two identical iron spheres, with the same radius. One is suspended over a support and one is hanged to a cable. Which will reach the higher temperature, if we give a heat Q to both spheres?
Knowing that the linear expansion follows this law: delta l= a*l*deltaTemp where "a" is constant and "l" is lenght, can you estimate the difference of temperature between the two spheres?

The problem is taken from the entrance test of 2/09/2014 to the Scuola Normale Superiore di Pisa.

Answer 1

I reasoned saying that the one above the support raises its center of mass so part of Q is tranformed in potential gravitational energy, but I'm not so convinced... Anyone has better ideas?

Answer 2

Ah, another interesting question - I will have to think about this one for a while : )

Answer 3

Ahah I've another one ready (always from the entrance test). In the meanwhile I go to sea so you've all the time, thanks (again) in advance!!

Answer 4

Ok - I need time, the brain is old.

Answer 5

I think you're on the right track, one sphere has to do work against gravity during its expansion, while the other gains a little bit of energy from gravity, due to the rise or fall of their centres of gravity.
The same Q is supplied to each sphere, so there will be a slight difference in the energy available to change the temperatures.
I worked out the numbers for spheres of 33kg mass (radius approx 0.1m) and imagined heating the spheres through 1000K to get an example value for Q.
The effect is extremely small.

Answer 6

Enough to pass the test, maybe :) . As soon as possible I'll write what was my answer. However I wanted to understand mainly if at least I was on the right way.

Answer 7

Yes, I think you are thinking the right physics.

Answer 8

I'm glad to listen that, after the test my idea seemed so crazy to me. So tomorrow I'm writing what I did and we'll check if our are the same. I wish you a nice day, see you tomorrow :)

Answer 9

On the left there's the heat given to the sphere attached to the rope; on the right the sphere on the support;
\[mC \Delta T_\max = mC \Delta T_\min+mg \Delta h\]
\[\Delta l= al \Delta T_\min\]
\[l=2R\]
\[\Delta l= 2 \Delta h\]

Substituting:
and simplifing m and 2:
\[C \Delta T_\max=C \Delta T_\min+gRa \Delta T_\min\]
now the problem asks for dt:
\[dt= \Delta T_\max-\Delta T_\min\]
and we can substitute Delta Tmax and Delta T min in the equation above.
Is it right?

Answer 10

I have for the ball that falls, mCdT=Q+mgaRdT and for the ball that rises, mCdT=Q-mgaRdT
where dT is the change in temperature (slightly different in each case)
then eliminating Q and rearranging gave\[\frac{ dThigh }{ dTlow }=\frac{ C-gaR }{ C+gaR }\]
Is it the same as your expression ? Probably, but I've not checked.

Answer 11

Yes is the same, thanks