What Is the Change in Entropy of a Resistor and the Universe?

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SUMMARY

The change in entropy of a 50-ohm resistor carrying a constant current of 1 A at a temperature of 27°C is zero, as the temperature remains constant and no heat is exchanged. The relevant equation for entropy change is dS = dQ / T, where dQ is the heat transfer. For the universe, the change in entropy can be calculated using the power generation formula, P = I²R, which indicates the energy dissipated per second. Understanding the relationship between resistance and heat transfer is crucial for calculating the total entropy change.

PREREQUISITES
  • Understanding of thermodynamics, specifically the concept of entropy.
  • Familiarity with electrical resistance and Ohm's Law.
  • Knowledge of heat transfer principles, particularly dQ = C dT.
  • Basic calculus for manipulating equations related to entropy.
NEXT STEPS
  • Study the principles of thermodynamics, focusing on entropy and heat transfer.
  • Learn about electrical power calculations, specifically P = I²R.
  • Explore the relationship between temperature, heat transfer, and entropy in closed systems.
  • Investigate the implications of constant temperature processes in thermodynamic systems.
USEFUL FOR

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

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Homework Statement



A 50-ohm resistor carrying a constant current of 1 A is kept at a constant temperature for 27 deg. C by a stream of cooling water. In a time interval of 1s, (a) What is the change in entropy of the resistor? (b) What is the change in entropy of the universe?


Homework Equations



dS = dQ / T
dQ = C dT


The Attempt at a Solution



For part a, I understand that the dQ of the resistor is zero because there is a constant temperature and current.

For part b, I do not understand the relationship between resistance (50 ohm), and dQ. I arrive at the correct answer by plugging in 300K (27 deg C) in for T, and 50 ohm in for dQ, but I would like to understand the correct relationship. Any help?
 
Physics news on Phys.org
Power generation (the amount of energy dissipated per second) is calculated as [itex]I^2R[/itex].
 

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