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Entropy involving current and resistance question 
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#1
Nov2811, 11:38 PM

P: 15

1. The problem statement, all variables and given/known data
A current of 10A is maintained for 1s in a resistor of 25Ω while the temperature of the resistor is kept constant at 27°C. This resistor is thermally insulated with a mass of 10g. If it has a specific heat of 836 J/kg*K. What is the entropy change of the resistor? Universe? 2. Relevant equations 3. The attempt at a solution Ok so if the wire is thermally insulated at a constant temperature of 27°C, Q=0 therefore W=0 because it is isothermal and adiabatic So if entropy is TdS= dU  dW so ΔS = 0J/K? However this is wrong...the answer is 5.78J/K and im confused at to how they got this... any suggestions? 


#2
Nov2911, 01:57 AM

HW Helper
Thanks
PF Gold
P: 4,851

I don't see how a resistor can have a current put thru it, be thermally insulated and maintained at constant temperature. Anybody?



#3
Nov2911, 01:04 PM

P: 15

Yea exactly, thats why I'm confused...being thermally insulated Q=0 and being held at a constant temperature it is isothermal. Therefore this is adiabatic and isothermal so...if the ΔS=ΔQ/T and there is no heat change well then there is no entropy change? Clearly I'm wrong though because the answers is 5.78 J/K.



#4
Nov2911, 02:03 PM

HW Helper
Thanks
PF Gold
P: 4,851

Entropy involving current and resistance question
I think we've been careless in reading the question.
Here's the situation: the resistor is packaged inside an insulator. The INSULATOR has c = 836 SI units and mass = 10g. The resistor clearly suffers no change in entropy because no heat flows to the resistor (the energy flowing to the resistor is work, as represented by a generator required to furnish the electricity). But the insulator does receive heat Q from the resistor, and will heat up (increase its temperature T) over the 1 sec time interval. So the insulator increases its entropy per ΔS = ∫dQ/T. If we ignore loss of heat to the environment, that will be the increase in entropy of the universe as well. T(t) of the insulator = T0 + (P/C)t where P = i^2*R, C = c*10g/1kg and t = time after starting the current. i = 10A, R = 25 ohms, and T(t=0) = T0 = 300K (assume insulator is at T=300K before current is applied). dQ(t)/dt for the insulator = CdT(t)/dt so dQ(t) = CdT(t) so now you can perform the integral. Careful of the difference between heat capacity C and specific heat c. C = 0.01c here since the insulation has mass = 10g. I'll try to come up with the actual answer while you work the problem and will let you know if I got the correct answer. 


#5
Nov2911, 02:19 PM

P: 15

Ah wonderful. Thanks for your help here now I get it. So the INSULATOR actually increases its entropy which is done by its temperature change which is T(t) = 300K + (10A^2*25Ω)/(8.36 J/K) = 599K
Now the Entropy change of the resistor inside the insulator is ΔS = ∫CdT/T = 8.36*ln(599K/300K) = 5.78 J/K Thank you rude man :P ! 


#6
Nov2911, 02:49 PM

HW Helper
Thanks
PF Gold
P: 4,851

So: ΔS of resistor = 0 ΔS of universe = 5.78J/K. Thanks for saving me the trouble of getting the answer. 


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