/*! 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 23 A sophomore with nothing bctter ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A sophomore with nothing bctter to do adds heat to \(0.350 \mathrm{~kg}\) of ice at \(0.0^{\circ} \mathrm{C}\) until it is all melted. (a) What is the change in entropy of the waler? (b) The source of heat is a very massive object at \(25.0^{\circ} \mathrm{C}\). What is the change in entropy of this object? (c) What is the total change in entropy of the water and the heat source?

Short Answer

Expert verified
The change in entropy of the water is \(0.428 \mathrm{kJK^{-1}}\), the change in entropy of the heat source is \(-0.392 \mathrm{kJK^{-1}}\), and the total change in entropy of the system is \(0.036 \mathrm{kJK^{-1}}\).

Step by step solution

01

Calculate the heat added to melt the ice

The heat \(Q\) required to melt the ice can be calculated using the equation \(Q = m \cdot L_f\), where \(m\) is the mass of the ice and \(L_f\) is the latent heat of fusion of ice (\(334 \mathrm{kJ/kg}\)). Substituting the given values gives \(Q = 0.350 \mathrm{kg} \times 334 \mathrm{kJ/kg} = 117 \mathrm{kJ}\).
02

Calculate the entropy change of the water

The change in entropy \(\Delta S\) of the water is given by \(\Delta S = Q / T\), where \(T\) is the absolute temperature in K. Since the ice was at \(0.0^{\circ}C\) or \(273.15\mathrm{K}\), the change in entropy is \(\Delta S = 117 \mathrm{kJ / 273.15 K} = 0.428 \mathrm{kJK^{-1}}\).
03

Calculate the entropy change of the heat source

The change in entropy of the heat source \(\Delta S_{source}\) can be calculated by \(\Delta S_{source} = -Q / T_{source}\). The temperature of the source is \(25.0^{\circ}C\) or \(298.15\mathrm{K}\), so the change in entropy of the source is \(\Delta S_{source} = -117 \mathrm{kJ / 298.15 K} = -0.392 \mathrm{kJK^{-1}}\).
04

Calculate the total change in entropy of the system

To calculate the total change in entropy \(\Delta S_{total}\) of the system, add the entropy change of the water and the entropy change of the source. Thus, \(\Delta S_{total} = \Delta S + \Delta S_{source} = 0.428 \mathrm{kJK^{-1}} - 0.392 \mathrm{kJK^{-1}} = 0.036 \mathrm{kJK^{-1}}\).

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 91Ó°ÊÓ!

Key Concepts

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

Entropy
Entropy is a fundamental concept in thermodynamics that refers to the measure of disorder or randomness in a system. When you think about melting ice, you're essentially thinking about a change in entropy. At 0°C, ice transforms into water as it absorbs heat, making the molecules within the substance increase their disorder.
This transformation from a solid to a liquid markedly increases entropy because the water molecules in liquid form have more freedom of movement.
  • Entropy (\( \Delta S \)) quantifies this change.
  • It is given by the formula: \( \Delta S = \frac{Q}{T} \), where \( Q \) is the heat added and \( T \) is the absolute temperature in Kelvin.
Considering the transformation of \( 0.350 \, \mathrm{kg} \) of ice at 0°C, the change in entropy of the water was calculated using the formula \( \Delta S = \frac{117 \, \mathrm{kJ}}{273.15 \, \mathrm{K}} = 0.428 \, \mathrm{kJ/K} \). This value showcases how energy and randomness increase upon melting.
Latent heat of fusion
The latent heat of fusion is the amount of heat required to convert a solid into a liquid at its melting point without changing its temperature. This is an important concept in determining how much energy is needed to melt ice, as it allows us to calculate the heat absorbed by the substance.The latent heat of fusion for ice is \( 334 \, \mathrm{kJ/kg} \). This means that every kilogram of ice requires \( 334 \, \mathrm{kJ} \) of energy to change its phase from solid to liquid without any temperature change. In our exercise:
  • The mass of ice is \( 0.350 \, \mathrm{kg} \).
  • The heat required can be calculated with the formula \( Q = m \times L_f \), which gives us \( Q = 0.350 \, \mathrm{kg} \times 334 \, \mathrm{kJ/kg} = 117 \, \mathrm{kJ} \).
This latent heat ensures that the ice will melt completely without raising its temperature, highlighting the energy stored in the process of melting.
Heat transfer
Heat transfer is the process of energy flowing from a warmer object to a cooler one. In our scenario, heat is transferred from a massive object at 25°C to the ice at 0°C, causing the ice to melt.Three key modes of heat transfer are conduction, convection, and radiation:
  • Conduction is the transfer of heat through a solid material, like the warmth moving from the surface of our heat source to the ice.
  • Convection involves the movement of fluid (liquid or gas) transferring heat.
  • Radiation transfers heat in the form of electromagnetic waves, such as light or infrared light.
In this case, conduction likely plays the most significant role, as heat flows from the heat source into the ice.
The entropy change for the heat source is calculated as:\( \Delta S_{source} = -\frac{Q}{T_{source}} \), where \( Q = 117 \, \mathrm{kJ} \) and \( T_{source} = 298.15 \, \mathrm{K} \). This results in an entropy change of \( -0.392 \, \mathrm{kJ/K} \), indicating heat lost by the source.

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

The Outo-cycle engine in a Mercedes-Benz SL.K230 has a compression ratio of 8.8 . (a) What is the ideal efficiency of the engine? Use \(\gamma=1.40 .\) (b) The engine in a Dodge Viper GT2 has a slightly higher compression ratio of 9.6 . How much increase in the ideal efficicncy results from this increase in the compression ratio?

The theorctical cfficicncy of an Otto cycle is \(63 \% .\) The ratio of the \(C_{p}\) and \(C_{V}\) heat capacities for the gas in the cngine is \(\frac{7}{5}\). (a) What is the compression ratio \(r ?\) (b) If the net work done per cycle is \(12.6 \mathrm{~kJ}\), how much heat is rejected and how much is absorbed by the engine during each cycle?

A Carnot engine whose high-temperature reservoir is at \(620 \mathrm{~K}\) takes in \(550 \mathrm{~J}\) of heat at this temperature in each cycle and gives up \(335 \mathrm{~J}\) to the low-temperature rescrvoir. (a) How much mechanical work does the cngine perform during cach cycle? What is (b) the temperature of the low-temperature reservoir; (c) the thermal cfficicncy of the cycle?

A cylinder contains oxygen at a pressure of \(2.00 \mathrm{~atm}\). The volume is \(4.00 \mathrm{I}_{2}\), and the temperature is \(300 \mathrm{~K}\). Assume that the oxygen may be treated as an ideal gas. The oxygen is carried through the following processes: (i) Heated at constant pressure from the initial state (state 1 ) to state 2 , which has \(T=450 \mathrm{~K}\). (ii) Cooled at constant volume to \(250 \mathrm{~K}\) (state 3 ). (iii) Compressed at constant temperature to a volume of \(4.00 \mathrm{~L}\) (state 4 ). (iv) Heated at constant volume to \(300 \mathrm{~K},\) which takes the system back to state 1 (a) Show these four processes in a \(p V\) -diagram, giving the numerical values of \(p\) and \(V\) in each of the four states. (b) Calculate \(Q\) and \(W\) for each of the four processes. (c) Calculate the net work done by the oxygen in the complete cycle. (d) What is the efficiency of this device as a heat engine? How does this compare to the efliciency of a Carnot-cycle engine operating between the same minimum and maximum temperatures of \(250 \mathrm{~K}\) and \(450 \mathrm{~K}\) ?

Entropy of Metabolism. An average sleeping person metabolizes at a rate of about \(80 \mathrm{~W}\) by digcsting food or burning fat. Typically, \(20 \%\) of this energy goes into bodily functions, such as cell repair, pumping blood, and other uses of mechanical energy, while the rest goes to heat. Most people get rid of all this excess heat by transferring it (by conduction and the flow of blood) to the surface of the body. where it is radiated away, The normal intemal temperature of the body (where the metabolism takes place) is \(37^{\circ} \mathrm{C},\) and the skin is typically \(7 \mathrm{C}^{3}\) cooler. By how much does the person's entropy change per sccond due to this heat transfer?

See all solutions

Recommended explanations on Physics Textbooks

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.