Change in entropy of a resistor

AI Thread Summary
The discussion focuses on calculating the temperature increase and entropy change of a thermally insulated resistor with given parameters. The power dissipated by the resistor is calculated using the formula P=I²R, leading to a temperature increase from the initial 20°C. There is a correction regarding the specific heat capacity, which should be in J/(kg-K) rather than J/(g-K), affecting the calculations. A participant suggests that the entropy change, initially calculated as ΔS = 0.273 J/K, may need reevaluation due to this unit discrepancy. Overall, the conversation emphasizes the importance of using correct units in thermodynamic calculations.
Elvis 123456789
Messages
158
Reaction score
6

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
    77.2 KB · Views: 948
Physics news on Phys.org
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.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top