A current of \(10 \mathrm{~A}\) is maintained for \(1 \mathrm{~s}\) in a resistor of \(25 \Omega\) while the temperature of the resistor is kept constant at \(27^{\circ} \mathrm{C}\). (a) What is the entropy change of the resistor? (b) What is the entropy change of the universe? The same current is maintained for the same time in the same resistor, but now thermally insulated, with the same initial temperature. If the resistor has a mass of \(10 \mathrm{~g}\) and a specific heat of \(836 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}:\) (c) What is the entropy change of the resistor? (d) What is the entropy change of the universe?

Entropy is a key concept in thermodynamics that measures the disorder or randomness of a system. It's often described as the amount of energy in a system that cannot do work. In practical terms, when energy is transferred or transformed, it can spread out and become less useful, increasing the system's entropy.
In our exercise, we witness entropy changes when energy is dissipated in a resistor. The formula for calculating entropy change is given by \( \Delta S = \frac{Q}{T} \), where \( Q \) is the heat transferred, and \( T \) is the temperature in Kelvins.
For example, when a current flows through a resistor and dissipates energy as heat while maintaining a constant temperature, the entropy of the resistor increases. In Part (a) of our exercise, this change in entropy was calculated to be 8.33 J/K for a temperature of 300 K.

Energy dissipation refers to the conversion of electrical energy into other forms such as heat or light. In the context of a resistor, electrical energy dissipates as heat due to the resistance. The amount of energy dissipated can be determined using the formula \( Q = I^2Rt \).
In our exercise, we calculated the energy dissipated in a 25 ohm resistor with a current of 10 amperes maintained for 1 second. Applying the formula, this yields \( Q = 10^2 \times 25 \times 1 = 2500 \text{J} \). This dissipated energy is responsible for the changes in entropy that follow.
Understanding how energy is transformed and dissipated helps to grasp why the entropy of a system changes as it absorbs and loses heat.

Specific heat is a property of a material that indicates how much heat energy is needed to raise the temperature of 1 kilogram of the material by 1 Kelvin. It plays an essential role in determining temperature changes in materials when they absorb or dissipate heat.
In Part (c) of our exercise, the specific heat value, denoted as \( c \), for the resistor's material is given as 836 J/kg·K. When the resistor is thermally insulated and electrically heated, we can calculate the temperature increase using \( \Delta T = \frac{Q}{mc} \). Here, \( Q \) is the heat added (2500 J), \( m \) is the mass (0.01 kg), and \( c \) is the specific heat.
Applying the values, \( \Delta T = \frac{2500}{0.01 \times 836} \text{K} \), results in a temperature increase of around 298.56 K. This specific heat value lets us determine how much the resistor's temperature rises, influencing the resultant entropy change.

Thermodynamics is the branch of physics that deals with heat, work, and the associated energy transformations. The laws of thermodynamics govern these processes and help us understand the natural tendencies of physical systems.
In this exercise, we explore concepts from the second law of thermodynamics, which states that the entropy of an isolated system always increases. For instance, in Part (d), the total entropy change of the universe when the resistor is thermally insulated is simply the entropy change of the resistor, since it is isolated from external heat exchanges.
Understanding thermodynamics laws allows us to predict how energy will behave in different scenarios. For instance, when the resistor's entropy increases by 5.77 J/K in an insulated setup, as calculated in our steps, we clearly observe the second law in action. Each part of the exercise ties back to these fundamental laws, showing their application in real-world problems.