Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 20: Problem 2

What is the magnitude of the change in entropy when \(6.00 \mathrm{~g}\) of steam at \(100{ }^{\circ} \mathrm{C}\) is condensed to water at \(100{ }^{\circ} \mathrm{C} ?\) a) \(46.6 \mathrm{~J} / \mathrm{K}\) c) \(36.3 \mathrm{~J} / \mathrm{K}\) b) \(52.4 \mathrm{~J} / \mathrm{K}\) d) \(34.2 \mathrm{~J} / \mathrm{K}\)

Short Answer

Expert verified
Answer: c) \(36.3 \mathrm{~J} / \mathrm{K}\)

Step by step solution

01

Identify the known values

We know the mass of steam (m) is 6.00 g, the temperature (T) in °C is 100°C, and the latent heat of vaporization for water (L) is 2260 J/g. First, we need to convert the temperature to Kelvin by adding 273.15 to 100°C: 100°C + 273.15 = 373.15 K.
02

Calculate the change in entropy

Using the formula for entropy change during a phase transition: ∆S = mL/T, we can calculate the change in entropy as follows: ∆S = (6.00 g)(2260 J/g) / (373.15 K)
03

Simplify and solve for ∆S

Simplify the expression and solve for ∆S: ∆S = 6.00 * 2260 / 373.15 ≈ 36.3 J/K
04

Compare the result to the given options

Our calculated change in entropy is approximately 36.3 J/K, which corresponds to option c. Therefore, the correct answer is: c) \(36.3 \mathrm{~J} / \mathrm{K}\)

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

Suppose a person metabolizes \(2000 .\) kcal/day. a) With a core body temperature of \(37.0^{\circ} \mathrm{C}\) and an ambient temperature of \(20.0^{\circ} \mathrm{C}\), what is the maximum (Carnot) efficiency with which the person can perform work? b) If the person could work with that efficiency, at what rate, in watts, would they have to shed waste heat to the surroundings? c) With a skin area of \(1.50 \mathrm{~m}^{2}\), a skin temperature of \(27.0^{\circ} \mathrm{C}\) and an effective emissivity of \(e=0.600,\) at what net rate does this person radiate heat to the \(20.0^{\circ} \mathrm{C}\) surroundings? d) The rest of the waste heat must be removed by evaporating water, either as perspiration or from the lungs At body temperature, the latent heat of vaporization of water is \(575 \mathrm{cal} / \mathrm{g}\). At what rate, in grams per hour, does this person lose water? e) Estimate the rate at which the person gains entropy. Assume that all the required evaporation of water takes place in the lungs, at the core body temperature of \(37.0^{\circ} \mathrm{C}\).

A refrigerator has a coefficient of performance of \(5.0 .\) If the refrigerator absorbs 40.0 cal of heat from the low temperature reservoir in each cycle, what is the amount of heat expelled into the high-temperature reservoir?

Why might a heat pump have an advantage over a space heater that converts electrical energy directly into thermal energy?

Find the net change in entropy when \(100 . \mathrm{g}\) of water at \(0^{\circ} \mathrm{C}\) is added to \(100 . \mathrm{g}\) of water at \(100 .{ }^{\circ} \mathrm{C}\)

It is desired to build a heat pump that has an output temperature of \(23^{\circ} \mathrm{C}\). Calculate the maximum coefficient of performance for the pump when the input source is (a) outdoor air on a cold winter day at \(-10.0^{\circ} \mathrm{C}\) and \((\mathrm{b})\) ground water at \(9.0^{\circ} \mathrm{C}\).
See all solutions

Recommended explanations on Physics Textbooks

Modern Physics

Read Explanation

Circular Motion and Gravitation

Read Explanation

Electrostatics

Read Explanation

Geometrical and Physical Optics

Read Explanation

Linear Momentum

Read Explanation

Kinematics Physics

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.

Sign-up for free