/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 27 20.27 A 15.0-kg block of ice at ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 20: Problem 27

20.27 A 15.0-kg block of ice at \(0.0^{\circ} \mathrm{C}\) melts to liquid water at \(0.0^{\circ} \mathrm{C}\) inside a large room that has a temperature of \(20.0^{\circ} \mathrm{C}\) . Treat the ice and the room as an isolated system, and assume that the room is large enough for its temperature change to be ignored. (a) Is the melting of the ice reversible or irreversible? Explain, using simple physical reasoning without resorting to any equations. (b) Calculate the net entropy change of the system during this process. Explain whether or not this result is consistent with your answer to part (a).

Short Answer

Expert verified
(a) Irreversible; (b) Net entropy change is +1261.95 J/K, confirming irreversibility.

Step by step solution

01

Understanding Reversibility

(a) For the process to be reversible, the transfer of heat between the ice and the room must occur at temperatures that are infinitesimally close to each other. Since the ice melts at \(0^{\circ} \mathrm{C}\) and the room is at a much higher temperature of \(20^{\circ} \mathrm{C}\), there is a significant temperature difference. Therefore, the melting process occurs rapidly, making it irreversible.
02

Calculating Entropy Change of the Ice

The entropy change for the ice can be calculated using the formula \(\Delta S = \frac{Q}{T}\), where \(Q\) is the heat absorbed and \(T\) is the temperature. For melting, \(Q = mL_f\) where \(m = 15.0\, \text{kg}\) and \(L_f = 334\, \text{J/g}\). Convert to Joules and use \(T = 273\, \text{K}\): \[Q = 15,000\, \text{g} \times 334\, \text{J/g} = 5,010,000\, \text{J}\]Calculate \(\Delta S_{\text{ice}} = \frac{5,010,000}{273} = 18,357.51\, \text{J/K}\).
03

Calculating Entropy Change of the Room

The entropy change for the room is negative, as it loses the same amount of energy the ice gains. Use the formula \(\Delta S = \frac{Q}{T}\), where \(T=293\, \text{K}\) for the room:\[\Delta S_{\text{room}} = -\frac{5,010,000}{293} = -17,095.56\, \text{J/K}\].
04

Finding Net Entropy Change

The net entropy change of the system is the sum of the entropy changes of the ice and the room:\[\Delta S_{\text{net}} = 18,357.51 + (-17,095.56) = 1,261.95\, \text{J/K}\].
05

Consistency with Irreversibility

(b) The net entropy change is positive, \(\Delta S_{\text{net}} = 1,261.95\, \text{J/K}\). This positive change is consistent with the second law of thermodynamics, confirming that the process is irreversible, as entropy has increased.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Reversible and Irreversible Processes
In thermodynamics, a process can be classified as either reversible or irreversible based on how the change occurs.
For a process to be reversible, it must happen infinitely slowly, allowing the system to be in equilibrium at each stage. There should be no loss of energy, such as heat, due to friction or other factors during the process.
In practical terms, a reversible process is an idealization, often not achievable in real scenarios because any finite temperature difference causes irreversible heat transfer.
  • A classic example of a reversible process is the slow compression or expansion of a gas within a piston that allows the gas to maintain thermal equilibrium with its surroundings.
  • An irreversible process, like the melting of ice in a warm room, occurs naturally without the opportunity to return to the original state without additional changes in external conditions.
In the given exercise, the rapid melting of the ice block due to a significant temperature difference from the surrounding room temperature makes it an irreversible process. The heat transfer is not slow enough for the process to be reversible, resulting in a net entropy change, consistent with real-world occurrences.
Second Law of Thermodynamics
The second law of thermodynamics is fundamental to understanding entropy and the behavior of physical systems.
It states that the total entropy of an isolated system can never decrease over time. In essence, natural processes tend to evolve towards a state of maximum entropy or disorder.
This law describes why certain processes are irreversible and dictates the direction of time in thermodynamic processes鈥攆rom low entropy to high entropy.
  • Mathematically, for any irreversible process, the entropy change delta S is always greater than zero for a closed system.
  • In reversible processes, the change in entropy, delta S, would be zero, denoting no net change in disorder.
In the context of the ice melting in a warm room, the second law is affirmed by the positive net change in entropy calculated in the exercise. This indicates the irreversibility of the process, as entropy increases systematically over time.
Entropy Calculation
Understanding the calculation of entropy change is key to mastering thermodynamic processes. Entropy, denoted as S, quantifies the amount of energy in a system unavailable to do work and is generally associated with disorder.
When calculating the change in entropy, deltaS, consider both the system gaining energy and the system losing energy.
The formula used for entropy change is given by: \[ \Delta S = \frac{Q}{T} \]where Q is the amount of heat transferred and T is the temperature at which the process takes place in Kelvin.
  • In the exercise, the melting ice (system gaining energy) results in positive entropy change since energy is absorbed to convert solid ice to liquid water.
  • The room (system losing energy) yields a negative entropy change, as it transfers heat to the ice.
To find the net entropy change, sum the entropy changes of the ice and the room, as shown in the exercise. The result reveals a positive net entropy, aligning with the fundamental principles of the second law, signifying an overall increase in disorder.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

20.36 . A lonely party balloon with a volume of 2.40 \(\mathrm{L}\) and containing 0.100 \(\mathrm{mol}\) of air is left behind to drift in the temporarily uninhabited and depressurized Intermational Space Station. Sunlight coming through a porthole heats and explodes the balloon, causing the air in it to undergo a free expansion into the empty station, whose total volume is 425 \(\mathrm{m}^{3}\) . Calculate the entropy change of the air during the expansion.

20.9. A refrigerator has a coefficient of performance of \(2.10 .\) In cach cycle it absorbs \(3.40 \times 10^{4} \mathrm{J}\) of heat from the cold reservoir. (a) How much mechanical energy is required each cycle to operate the refrigerator? (b) During each cycle, how much heat is discarded to the high-temperature reservoir?

20.34 . A box is separated by a partition into two parts of equal volume. The left side of the box contains 500 molecules of nitrogen gas; the right side contains 100 molecules of oxygen gas. The two gases are at the same temperature. The partition is punctured, and equilibrium is eventually attained. Assume that the volume of the box is large enough for each gas to undergo a free expansion and not change temperature, (a) On average, how many molecules of each type will there be in either half of the box? (b) What is the change in entropy of the system when the partition is punctured? the (c) What is the probability that the molecules will be found in the same distribution as they were before the partition was punctured - that is, 500 nitrogen molecules in the left half and 100 oxygen molecules in the right half?

20.45. An experimental power plant at the Natural Energy Laboratory of Hawaii generates electricity from the temperature gradient of the ocean. The surface and deep-water temperatures are \(27^{\circ} \mathrm{C}\) and \(6^{\circ} \mathrm{C}\) , respectively. (a) What is the maximum theoretical effciency of this power plant? (b) If the power plant is to produce 210 \(\mathrm{kW}\) of power, at what rate must heat be extracted from the warm water? At what rate must heat be absorbed by the cold water? Assume the maximum theoretical efficiency. (c) The cold water that enters the plant leaves it at a temperature of \(10^{\circ} \mathrm{C}\) . What must be the flow rate of cold water through the system? Give your answer in \(\mathrm{kg} / \mathrm{h}\) and \(\mathrm{L} / \mathrm{h}\) .

20\. 07. What compression ratio \(r\) must an Otto cycle have to achieve an ideal efficiency of 65.0\(\%\) if \(\gamma=1.40 ?\)
See all solutions

Recommended explanations on Physics Textbooks

Fundamentals of Physics

Read Explanation

Measurements

Read Explanation

Solid State Physics

Read Explanation

Electricity

Read Explanation

Kinematics Physics

Read Explanation

Mechanics and Materials

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.