How Does Current Flow Affect Entropy in a Resistor and the Universe?

[Brewer]

Homework Statement

A 50 ohm resistor carries a current of 250 mA and is maintained at a constant temperature of 303K by thermal contact with a heat sink.

Calculate the entopy change of

i) the resistor, and
ii) the universe,

if the current flows for 30 seconds.

Homework Equations

dS >= 0?

The Attempt at a Solution

I'm not sure how to attempt this question for a physical problem. Up until now I've been restricted to doing questions of this type based upon ideal gases, so I'm not sure how I would apply that to a physical system like this. Any hints anyone?

 

[Brewer]

Right, I've had another look at this.

I've said that dQ=dW and that work = power * time, so dW = Pdt = I*I*R*dt

From here I've substituted this into the equation $$dS = \int \frac{I^2R}{T}dt$$ with limits 0 and 3.

Following this through I get an answer of 0.031 J/kg. Does this sound about right? Then for the entropy change of the universe it'll be just the negative of this answer?

 

[m2003]

I would approach this problem by first understanding the concept of entropy and how it applies to physical systems. Entropy is a measure of the disorder or randomness in a system and is related to the amount of energy that is unavailable for work. In this case, the resistor is a physical system that is connected to a heat sink, which means that there is a transfer of energy (in the form of heat) between the two systems.

To calculate the entropy change of the resistor, we can use the formula dS = dq/T, where dq is the amount of heat transferred and T is the temperature. In this case, we know that the resistor is maintained at a constant temperature of 303K and there is a current of 250 mA flowing through it for 30 seconds. This means that the amount of heat transferred is given by dq = I^2Rdt, where I is the current, R is the resistance, and dt is the time. Plugging in the values, we get dq = (0.25A)^2 * 50ohm * 30s = 0.375 J.

Substituting this into the formula, we get dS = 0.375 J / 303K = 0.0012 J/K. This is the change in entropy of the resistor due to the flow of current for 30 seconds.

To calculate the entropy change of the universe, we need to consider the entire system, which includes the resistor, the heat sink, and the surroundings. In this case, the heat sink is maintained at a constant temperature, so there is no change in its entropy. However, the surroundings will experience a change in entropy due to the transfer of heat from the resistor. Using the same formula as before, we can calculate the change in entropy of the surroundings, which will be equal and opposite to the change in entropy of the resistor.

Therefore, the change in entropy of the universe will be zero, as the increase in entropy of the surroundings will cancel out the decrease in entropy of the resistor. This is in line with the second law of thermodynamics, which states that the total entropy of an isolated system (such as the universe) will always increase or remain constant.