How Does Current Flow Affect Entropy in a Resistor and the Universe?

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SUMMARY

The discussion focuses on calculating the entropy change of a 50 ohm resistor carrying a current of 250 mA at a constant temperature of 303K over a duration of 30 seconds. The participant correctly identifies that the work done can be expressed as dW = I²Rdt, leading to the integration of dS = ∫(I²R/T)dt. The calculated entropy change for the resistor is 0.031 J/kg, and the entropy change for the universe is the negative of this value, confirming the second law of thermodynamics.

PREREQUISITES
  • Understanding of thermodynamics principles, specifically entropy.
  • Familiarity with electrical concepts such as current, resistance, and power.
  • Knowledge of calculus for integration in physical equations.
  • Basic understanding of heat transfer and thermal equilibrium.
NEXT STEPS
  • Study the second law of thermodynamics and its implications on entropy.
  • Learn about Joule heating and its effect on resistors in thermal systems.
  • Explore the relationship between electrical energy and thermal energy in resistive materials.
  • Investigate the concept of entropy in different physical systems beyond ideal gases.
USEFUL FOR

Students in physics or engineering, particularly those studying thermodynamics and electrical systems, as well as anyone interested in the relationship between current flow and entropy changes in physical systems.

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Homework Statement


A 50 ohm resistor carries a current of 250 mA and is maintained at a constant temperature of 303K by thermal contact with a heat sink.

Calculate the entopy change of

i) the resistor, and
ii) the universe,

if the current flows for 30 seconds.


Homework Equations


dS >= 0?


The Attempt at a Solution


I'm not sure how to attempt this question for a physical problem. Up until now I've been restricted to doing questions of this type based upon ideal gases, so I'm not sure how I would apply that to a physical system like this. Any hints anyone?
 
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Right, I've had another look at this.

I've said that dQ=dW and that work = power * time, so dW = Pdt = I*I*R*dt

From here I've substituted this into the equation [tex]dS = \int \frac{I^2R}{T}dt[/tex] with limits 0 and 3.

Following this through I get an answer of 0.031 J/kg. Does this sound about right? Then for the entropy change of the universe it'll be just the negative of this answer?
 

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