Large heat reservoirs are available at 900 K(H) and 300 K(C). (a) 100 cal of heat are removed

step 1 Hello and welcome to Digital Tea with Mr. E, where we will see what's brewing in the world of ph

Large heat reservoirs are available at 900 K(H) and 300...

Large heat reservoirs are available at $900 \mathrm{~K}(H)$ and $300 \mathrm{~K}(C)$. (a) 100 cal of heat are removed from the reservoir $H$ and added to $C$. What is the entropy change of the universe? (b) A reversible heat engine operates between $H$ and $C$. For each 100 cal of heat removed from $H$, what work is done and what heat is added to $C$ ? (c) What is the entropy change of the universe in the process of part (b) above? (d) A real heat engine is operated as a heat pump removing heat from $C$ and adding heat to $H$. What can be said about the entropy change in the universe produced by the heat pump?

Large heat reservoirs are available at $900 \mathrm{~K}(H)$ and $300 \mathrm{~K}(C)$.
(a) 100 cal of heat are removed from the reservoir $H$ and added to $C$. What is the entropy change of the universe?
(b) A reversible heat engine operates between $H$ and $C$. For each 100 cal of heat removed from $H$, what work is done and what heat is added to $C$ ?
(c) What is the entropy change of the universe in the process of part
(b) above?
(d) A real heat engine is operated as a heat pump removing heat from $C$ and adding heat to $H$. What can be said about the entropy change in the universe produced by the heat pump?

Key Concepts

Heat Engines

Heat engines are devices that convert thermal energy into mechanical work by operating between two reservoirs at different temperatures. Their performance is governed by the principles of energy conversion and the limitations imposed by the Second Law of Thermodynamics, with reversible engines serving as the model for maximum efficiency.

Heat Reservoirs

Heat reservoirs are idealized systems with very large thermal capacities that can absorb or release heat without undergoing a significant change in temperature. They serve as the source and sink in thermodynamic processes, making them critical to the analysis of heat engines, heat pumps, and entropy changes in the universe.

Carnot Efficiency

Carnot efficiency represents the theoretical maximum efficiency that any heat engine operating between two heat reservoirs can achieve. It is derived from the temperatures of the hot and cold reservoirs and provides a standard to compare the performance of real engines by highlighting how irreversibilities reduce efficiency in practical applications.

Heat Pumps

Heat pumps operate by transferring heat from a colder region to a hotter region, against the natural flow of heat, typically requiring an input of work. When analyzing heat pumps, considerations of entropy production become essential, as real devices produce additional entropy compared to their ideal, reversible counterparts.

Second Law of Thermodynamics

The Second Law of Thermodynamics states that the total entropy of an isolated system can never decrease over time, with irreversible processes always resulting in an increase in universe entropy. This concept provides the foundation for analyzing the feasibility and spontaneity of thermodynamic processes, such as the heat transfer between reservoirs or the operation of heat engines and pumps.

Entropy

Entropy is a thermodynamic property that quantifies the amount of disorder or randomness in a system. Changes in entropy during a process help determine whether a process is reversible or irreversible and play a crucial role in understanding the overall direction of natural processes, as outlined by the Second Law of Thermodynamics.

Reversible Processes

Reversible processes are idealized thermodynamic processes that occur infinitely slowly, allowing the system to remain in thermodynamic equilibrium at all times. In such processes, the total entropy change of the universe is zero, serving as an important benchmark for the maximum possible efficiency of real processes and devices.

Transcript

00:01 Hello and welcome to digital tea with mr.

00:03 E, where we will see what's brewing in the world of physics solutions.

00:07 So in this particular problem, we have a heat engine that is operating between a high temperature reservoir of 545 kelvin and a low temperature of 325 kelvin.

00:19 And we also know that the heat flowing out of the heat engine, which we will call qh, is equal to 50.

00:30 Joules and some of it is diverted into work and the rest of it is diverted into heat that is exhausted into the low temperature reservoir, which we know to be 40 joules.

00:46 So not that the question asks for it, but for overall knowledge, you should be able to determine the amount of work done by this heat engine per cycle or so.

01:02 Simply by stating that the qh should equal the work done plus the ql.

01:13 So we can see that the work in this particular case would this be the subtraction of qh and ql, which would be there's 10 joules of work done.

01:21 Alas, this particular problem is asking us to analyze the entropy of the whole system.

01:27 So when you're solving for entropy, you need to do a separate entropy calculation for each part of the system.

01:36 And by each part, i mean an entropy calculation for the high -temperature reservoir and a separate high -tempathy calculation for the low -temperature reservoir.

01:45 And the second law of thermodynamics predicts that when you consider the entire entropy change of the system, it should always come out to be a positive number.

01:54 In other words, the universe will always be more disordered because of the operation of a heat engine then it would have been had the heat engine never existed.

02:05 So let's take a look at the calculations and see if it matches the theory.

02:09 So the entropy calculation for the high temperature reservoir will be equal to qh over t .h.

02:24 And the entropy calculation for the low temperature reservoir will be equal to ql over tl.

02:34 So, in our calculators, let's type in the qh of 50 and divide it by the th of 545.

02:45 And when you do that, you get 0 .092 joules per kelvin.

02:57 Now, you have to decide if that's a positive or negative entropy change.

03:02 And it's always a negative entropy change if heat is, being removed from something or if the temperature is in a sense dropping.

03:12 So in this case we have heat removing being removed so this is a negative entropy change.

03:18 And then on the other side we have delta s for the low temperature and that would simply be the ql which is 40 and dividing it by the tl which is 325 and when you run those numbers you get a grand total of 0 .123.

03:37 Joules per kelvin.

03:40 And then define the change of entropy for the whole system...

Please give Ace some feedback