Change in entropy of a resistor

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Homework Help Overview

The discussion revolves around the change in entropy of a thermally insulated resistor, given specific parameters such as resistance, mass, specific heat capacity, and current. The problem involves calculating the temperature increase and the corresponding change in entropy as current flows through the resistor.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the calculations related to the mass of the resistor and the specific heat capacity, questioning the values used in the original poster's solution. There is also speculation about the nature of part (b) being a trick question.

Discussion Status

Several participants have provided feedback on the calculations, particularly regarding the mass and specific heat capacity. There is an ongoing exploration of the assumptions made in the problem setup, and some participants are awaiting further responses before drawing conclusions.

Contextual Notes

There are noted discrepancies in the mass used in calculations and the units for specific heat capacity, which are under discussion. The original poster expresses uncertainty about their solution, indicating a need for clarification on these points.

Elvis 123456789
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Homework Statement


Consider a thermally insulated resistor with resistance R=20 Ω and mass m=5.0 g. The resistor is made of a material with specific heat capacity c=850 J/(g-K) and carries a current of 2.0 A for a time period of 1.0 s.

a) Calculate the increase in the temperature of the resistor if it has initial temperature Tin=20oC. Recall that the power dissipated by a resistor is given by P=I²R.

b) Determine the increase in the resistor's entropy over the period that the current passes through it.

Homework Equations

The Attempt at a Solution


My solution is in the attachment. I just wanted somebody to have a look at it and let me know if it looks okay since I don't feel too sure about it.
 

Attachments

  • thermo hw.png
    thermo hw.png
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I note that you used 50g for your mass in the final calculation rather than 5g...
 
gneill said:
I note that you used 50g for your mass in the final calculation rather than 5g...
oh woops, the answer should have been ΔS = 0.273 J/K
 
I think part (b) is a trick question but I'm going to wait for other responses if any.
As a hint I don't agree with the computed answer in post 3.
 
Your specific heat capacity is unrealistic. It should be per kilo not gram.
 
alchemistoff said:
Your specific heat capacity is unrealistic. It should be per kilo not gram.
Yes, the approach is correct, but the heat capacity is off by a factor of 1000.
 

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