Entropy Change Q: 10 Ohm Resistor @ 300K, 5A for 2 Mins

AI Thread Summary
A 10 Ohm resistor at 300K with a current of 5A for 2 minutes generates heat, leading to entropy changes in both the resistor and the universe. The power dissipated can be calculated using the formula I^2R, which equals 250W. Since the process is irreversible, the change in entropy for the resistor can be determined by considering a reversible path between its initial and final states. Additionally, the entropy change for the universe must account for the surroundings' temperature, which is necessary for accurate calculations. Overall, the total entropy of the universe increases due to the irreversible nature of the process.
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Homework Statement



A 10 Ohm resistor is held at a temp of 300K. A current of 5A is passed through the resistor for 2 mins. Ignoring changes in the source of current, what is the change of entropy in:
(a) the resistor
(b) the universe


Homework Equations





The Attempt at a Solution



Ok so not sure how to do this..

Obviously i know how to work out the total power dissipated (I^2 R)..

dS = dQ rev/T

now not sure what to do... the process isn't reversible right?


im guessing the total entropy of the universe increases..?

help please! :)
 
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bon said:

Homework Statement



A 10 Ohm resistor is held at a temp of 300K. A current of 5A is passed through the resistor for 2 mins. Ignoring changes in the source of current, what is the change of entropy in:
(a) the resistor
(b) the universe
What is the temperature of the surroundings?

The overall process is not reversible. But entropy is calculated for the resistor and surroundings separately. Determine the reversible path between the initial and final state for the resistor and calculate the change in entropy for the resistor. Then find the change in entropy for the reversible path between the initial and final states for the surroundings. You will need the temperature of the surroundings though.

AM
 

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