Entropy of running water around a resistor

AI Thread Summary
The discussion revolves around calculating the entropy changes in a resistor, running water, and the universe when a 10A current flows through a 20-ohm resistor at a constant temperature of 10°C. The participant notes that the resistor's entropy remains unchanged since the energy is work done rather than heat transfer. They calculate the power dissipated by the resistor as 2000W, equating to 2000J. The main challenge lies in correctly applying the entropy change formula while considering the constant temperature and converting temperature to Kelvin. Ultimately, they conclude that the entropy change for the universe is equivalent to that of the running water.
bmarson123
Messages
24
Reaction score
0

Homework Statement


An electric current of 10A runs for 1 minute through a 20 ohm Resistor which is maintained at 10 degC by being immersed in running water. What are the entropy changes in a) the resistor b) the running water c) the universe


Homework Equations



P = I2R

dS = dQ/T

The Attempt at a Solution



For a) there is no change in the entropy as it is work done by the resistor, not heat.

b) I've calculated P = 2000W = 2000J

The problem I'm having is formulating the path between the initial and final states of the system because of the constant temperature. I'm also not sure if I'm using the right equation?

c) This will be the same as the answer for b) I think.
 
Physics news on Phys.org
Figured it out!

Forgot to convert the temperature to kelvin and then got everything written down in a mess!
 
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