Thermodynamics - shaking a bottle with liquid

AI Thread Summary
Shaking a closed bottle containing liquid increases the temperature due to internal viscous forces, with energy transferred from the shaker to the liquid. Although no heat is transferred into the liquid, work is done on it, as indicated by the first law of thermodynamics. The kinetic energy imparted to the liquid is dissipated through viscous flow, converting it into heat energy. This process is analogous to the Joule paddle experiment, where mechanical work results in a temperature rise without external heat flow. Ultimately, the work done on the liquid manifests as an increase in thermal energy.
zorro
Messages
1,378
Reaction score
0
A closed bottle contains a liquid. The bottle is shaken vigorously for 5 minutes. It is found that the temperature of the liquid is increased. Is heat transferred to the liquid? Is work done on the liquid. Neglect expansion on heating.

I think the temperature increases due to the internal viscous force of the liquid. Heat is not transferred to the liquid as the bottle is closed and no work is done the the liquid because change in volume is 0.
 
Physics news on Phys.org


Shaking the bottle transfers energy (from the person or thing shaking) to the contents.
 


Do you mean the muscular energy of the person (shaking) is converted into kinetic energy of the liquid and consequently heat energy? So heat is transferred and work is done on the liquid.
Is it right?
 


Kinetic energy is definitely imparted to the fluid, but no work is done (assuming that at the end, the fluid is at rest and has not been raised or lowered). The kinetic energy is dissipated by viscous flow and converted into heat energy.
 
Andy Resnick said:
Kinetic energy is definitely imparted to the fluid, but no work is done (assuming that at the end, the fluid is at rest and has not been raised or lowered). The kinetic energy is dissipated by viscous flow and converted into heat energy.
Yes. It is a bit tricky. There is no heat flow into the liquid yet its temperature increases. From the first law: dQ = dU + dW so if dQ = 0 then dW = -dU. So work is done on the liquid. It is just that the work is converted to heat.

This is similar to the Joule paddle experiment.

AM
 


Andrew Mason said:
There is no heat flow into the liquid yet its temperature increases. From the first law: dQ = dU + dW so if dQ = 0 then dW = -dU. So work is done on the liquid. It is just that the work is converted to heat.

This is similar to the Joule paddle experiment.

AM

Thankyou!
 
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...

Similar threads

Back
Top