# Entropy involving current and resistance question

by jrklx250s
Tags: current, entropy, involving, resistance
 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?
 HW Helper Thanks PF Gold P: 4,914 I don't see how a resistor can have a current put thru it, be thermally insulated and maintained at constant temperature. Anybody?
 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.
 HW Helper Thanks PF Gold P: 4,914 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.
 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 !
HW Helper
Thanks
PF Gold
P: 4,914
 Quote by jrklx250s 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 !
No, the resistor entropy does not change, because no heat flowed into it (I know it sounds goofy, but you have to interpret "heat" very strictly as that energy which flows by virtue of a difference in temperature. The resistor receives no heat flow; its increase in internal energy comes from the work done by, for example, an external generator used to supply its electrical energy.

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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