Entropy of running water around a resistor

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SUMMARY

The discussion focuses on calculating the entropy changes associated with a 20 ohm resistor carrying a 10A electric current for 1 minute while maintained at 10°C in running water. The power dissipated by the resistor is calculated as 2000W, leading to an energy transfer of 2000J. The entropy change for the resistor is determined to be zero since the work done does not involve heat transfer. The entropy change for the running water is equivalent to that of the resistor, with the final calculations corrected by converting the temperature to Kelvin.

PREREQUISITES
  • Understanding of electrical power calculations (P = I²R)
  • Knowledge of thermodynamic entropy equations (dS = dQ/T)
  • Familiarity with temperature conversion from Celsius to Kelvin
  • Basic concepts of heat transfer in thermodynamic systems
NEXT STEPS
  • Study the principles of thermodynamic entropy in closed systems
  • Learn about heat transfer mechanisms in resistive materials
  • Explore the implications of constant temperature processes in thermodynamics
  • Investigate the relationship between electrical energy and thermal energy in resistors
USEFUL FOR

Students in physics or engineering, particularly those studying thermodynamics and electrical circuits, as well as educators looking for practical examples of entropy calculations in resistive systems.

bmarson123
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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!
 

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