Does Copper's Lack of Dipole Movement Increase Thermal Radiation in Calorimetry?

AI Thread Summary
The discussion centers on the thermal radiation loss from oil at 80 degrees Celsius in a copper calorimetry cup. Participants debate whether this heat loss is significant, with some suggesting it is negligible due to copper's excellent thermal conductivity and low heat capacity. The lack of dipole movement in copper is mentioned as a factor that may influence thermal radiation, but the consensus leans towards minimal impact on calorimetry values. The Stefan-Boltzmann Law is referenced to understand the relationship between temperature and radiation heat transport. Overall, the heat loss by thermal radiation in this scenario is considered to be low and unlikely to affect the measurements significantly.
nithin
Messages
29
Reaction score
0
I have a question. Assuming I have oil at 80 degrees in a copper calorimetry cup , will the heat loss by thermal radiation be very high and will it affect my values?

I am thinking that it would not and that it would be negligible. But i do not know why. I heard somewhere that it would be high due to copper not having a dipole movement. Can someone please explain to me why?
 
Physics news on Phys.org
Your own experience with hot drinks surely tells you that a liquid at 80C cools pretty quickly whatever its in.

Copper is a very good conductor of heat and has a low heat capacity so it won't help much to keep the oil warm.

The heat capacity of a substance depends on its internal degrees of freedom for vibration, that's where the atomic dipole movement comes in.
 
nithin said:
I have a question. Assuming I have oil at 80 degrees in a copper calorimetry cup, will the heat loss by thermal radiation be very high and will it affect my values?

I am thinking that it would not and that it would be negligible. But i do not know why. I heard somewhere that it would be high due to copper not having a dipole movement. Can someone please explain to me why?
See the discussion of radiation heat transport and the Stefan-Boltzmann Law
http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/stefan.html

Let Thot = 80°C or 353 K and Tcold = 25° or 298 K. One must use absolute temperature in the S-B law.
 
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...
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...
Back
Top