Question

One end of a metal rod is in contact with a thermal reservoir at 700. \(K\), and the other end is in contact with a thermal reservoir at \(100 .\) K. The rod and reservoirs make up an isolated system. If \(8500 .\) J are conductec from one end of the rod to the other uniformly (no change in temperature along the rod) what is the change in entropy of (a) each reservoir, (b) the rod, and (c) the system?

Short answer

Question: Calculate the change in entropy for (a) the cold reservoir, (b) the hot reservoir, and (c) the entire system for a metal rod in contact with two thermal reservoirs at different temperatures transferring 8500 J of heat. Answer: (a) 85 J/K, (b) -12.14 J/K, (c) 72.86 J/K.

Step-by-step solution

Heat transfer to the cold reservoir

The heat transferred to the cold reservoir will be equivalent to the heat transferred through the rod, which is 8500 J.

Calculate the change in entropy for the cold reservoir

Now, we can use the formula for change in entropy to find the entropy change for the cold reservoir: ∆S_cold = Q/T_cold ∆S_cold = 8500 J / 100 K ∆S_cold = 85 J/K So, the change in entropy for the cold reservoir is 85 J/K.

Heat transfer from the hot reservoir

The heat transferred from the hot reservoir will be equivalent to the heat transferred through the rod, so it will also be 8500 J.

Calculate the change in entropy for the hot reservoir

Now, we can use the formula for change in entropy to find the entropy change for the hot reservoir: ∆S_hot = -Q/T_hot ∆S_hot = -8500 J / 700 K ∆S_hot = -12.14 J/K The negative sign indicates that the entropy of the hot reservoir decreases. So, the change in entropy for the hot reservoir is -12.14 J/K.

Calculate the change in entropy for the entire system

To find the change in entropy for the entire system, we can simply add the entropy changes for both reservoirs: ∆S_system = ∆S_hot + ∆S_cold ∆S_system = -12.14 J/K + 85 J/K ∆S_system = 72.86 J/K So, the change in entropy for the entire system is 72.86 J/K. Finally, we have the change in entropy for (a) the cold reservoir: 85 J/K, (b) the hot reservoir: -12.14 J/K, and (c) the entire system: 72.86 J/K.

Key concepts

Thermal Reservoir

A thermal reservoir, sometimes also referred to as a heat reservoir, is an entity with a thermal capacity that is effectively infinite compared to the system of interest. What this means is that the temperature of the thermal reservoir does not change significantly when an amount of heat is added or removed.

For instance, in the textbook exercise, one end of the metal rod is in contact with a thermal reservoir at 700 K, and the other end with a reservoir at 100 K. Despite heat being transferred from the hot to the cold reservoir through the rod, the reservoirs maintain their respective temperatures due to their vast capacity to absorb or supply energy without notable temperature changes. In real-world scenarios, large bodies of water or the atmosphere itself can act as thermal reservoirs.

Heat Transfer

Heat transfer is the process of thermal energy moving from one body or substance to another with a different temperature. The second law of thermodynamics explains that heat naturally flows from a warmer object to a cooler one.

There are three main modes of heat transfer: conduction, convection, and radiation. In the given problem, heat transfer occurs through conduction—the direct transfer of thermal energy through a material, from the high-temperature end of the rod (the hot reservoir) to the low-temperature end (the cold reservoir). Conduction is particularly efficient in metals, which are typically good conductors of heat.

Isolated System

An isolated system is a physical system that doesn't exchange any matter or energy with its surroundings. Consequently, the total energy and mass within an isolated system remain constant over time.

In the exercise, the metal rod and both reservoirs are described as an isolated system, which means that the heat transfer (8500 J) is contained within the rod and the reservoirs. No external heat flow or mass transfer occurs in or out of this system. Therefore, while the internal distribution of energy changes, the overall energy of the system remains the same, and its total entropy is not influenced by external factors.

Entropy Formula

Entropy, a central concept in thermodynamics, quantifies the level of disorder or randomness within a system. The greater the disorder, the higher the entropy. To calculate the change in entropy (\( \triangle S \(\3\)) within a system where heat transfer occurs, one can use the entropy formula:
\( \triangle S = \frac{Q}{T} \(\3\)), where \( Q \(\3\)) is the heat added to or removed from a system and \( T \(\3\)) is the temperature. It's crucial to maintain the units correctly—heat in joules and temperature in kelvin for the SI system.

Applying this concept to our exercise, the varying entropy values for the hot and cold reservoirs result from their temperatures. For the cold reservoir at 100 K, adding 8500 J of heat increased its entropy, while for the hot reservoir at 700 K, losing the same amount of heat meant a decrease in its entropy. The overall system entropy change also shows the net result of these heat transfers, reflecting the second law of thermodynamics: the total entropy of an isolated system can never decrease.