/*! 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 28 At very low temperatures, the mo... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

At very low temperatures, the molar specific heat \(C_V\) of many solids is approximately \(C_V=A T^3\), where \(A\) depends on the particular substance. For aluminum, \(A=3.15 \times 10^{-5} \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K}^4\). Find the entropy change for \(4.00 \mathrm{~mol}\) of aluminum when its temperature is decreased from \(8.00 \mathrm{~K}\) to \(5.00 \mathrm{~K}\).

Short Answer

Expert verified
The entropy change is approximately -0.0163 J/K.

Step by step solution

01

Understanding Entropy Change Formula

Entropy change can be calculated using the formula for reversible processes in thermodynamics, \( \Delta S = \int \frac{dQ}{T} \), where \( dQ = nC_VdT \). Here, \( Q \) is the heat added or removed, \( n \) is the number of moles, and \( C_V \) is the specific heat at constant volume.
02

Express Specific Heat in Terms of Temperature

Given \( C_V = A T^3 \), substitute \( C_V \) in the equation for \( dQ \) such that \( dQ = nA T^3 dT \).
03

Setup Numerical Integration Bounds

Set up the integral for entropy change from initial temperature \( T_1 = 8.00 \) K to final temperature \( T_2 = 5.00 \) K: \[ \Delta S = nA \int_{T_1}^{T_2} T^2 dT \].
04

Perform Integration of the Function

Integrate \( T^2 \):\[ \int T^2 dT = \frac{T^3}{3} + C \].
05

Evaluate the Definite Integral

Apply the definite integral from \( T_1 = 8.00 \) K to \( T_2 = 5.00 \) K:\[ \left[ \frac{T^3}{3} \right]_{8}^{5} = \frac{5^3}{3} - \frac{8^3}{3} = \frac{125}{3} - \frac{512}{3} = -129 \].
06

Calculate Entropy Change

Plug \( n = 4.00 \) mol and \( A = 3.15 \times 10^{-5} \) J/mol K^4 into the calculated integral:\[ \Delta S = 4.00 \times 3.15 \times 10^{-5} \times (-129) \approx -0.0163 \text{ J/K} \].

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.

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 Change
Entropy change is a fundamental concept in thermodynamics, representing the degree of disorder or randomness within a system. Simply put, it explains how energy disperses in a system.
In mathematical terms, entropy change for a reversible process is given by the formula \( \Delta S = \int \frac{dQ}{T} \). Here, \( dQ \) is the infinitesimal amount of heat added or removed from the system, and \( T \) is the absolute temperature.
It's important because it helps us understand if a process is spontaneous or if it can occur naturally. In the context of the given exercise, we are calculating how the entropy for 4 moles of aluminum changes when its temperature decreases from 8 K to 5 K. Knowing the change helps predict how the material behaves at low temperatures.
Specific Heat
Specific heat is a crucial property of materials in thermodynamics, defining how much heat a material can absorb without undergoing a significant temperature change. More technically, it is the amount of heat needed to raise the temperature of one gram of a substance by one degree Celsius.
In our exercise, the specific heat \( C_V \) is expressed as \( A T^3 \), exhibiting temperature dependence, especially noticeable at low temperatures. This temperature dependence is useful in accurately modeling the behavior of materials such as aluminum.
By understanding specific heat, one can predict how a substance will respond to external heat, which is essential in applications ranging from household heating to industrial material design.
Temperature Dependence
Temperature dependence is a sensitive factor in thermodynamics; it describes how different physical properties change as temperature changes. As the temperature of a material changes, its properties like specific heat can vary significantly.
In this exercise, the specific heat \( C_V \) of aluminum is dependent on the cube of its temperature, \( C_V = A T^3 \). This behavior is common in solids at low temperatures as per the Debye model, which tries to explain how specific heat changes with temperature at low values, often deviating from classical models.
Understanding temperature dependence is vital for calculating heat-related phenomena, as it allows for precise adjustments in heating and cooling processes, especially in scientific and engineering applications where exact temperature control is crucial.
Molar Specific Heat
Molar specific heat extends the concept of specific heat to multiple moles of a substance. It measures how much heat energy is needed to raise the temperature of one mole of a substance by one Kelvin.
In the given exercise, we are dealing with molar specific heat in terms of \( C_V = A T^3 \). The constant \( A \) provided (for aluminum, \( A = 3.15 \times 10^{-5} \text{ J/mol} \cdot \text{K}^4 \)) gives insight into how aluminum's heat capacity transforms at low temperatures.
Molar specific heat is integral to determine energy changes in reactions involving gases, liquids, and solids. It also plays a key role in designing systems that must manage heat efficiently, ensuring they do not overheat or lose energy unnecessarily.

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

A heat pump is used to heat a building. The external temperature is less than the internal temperature. The pump's coefficient of performance is \(3.30\), and the heat pump delivers \(7.54 \mathrm{MJ}\) as heat to the building each hour. If the heat pump is a Carnot engine working in reverse, at what rate must work be done to run it?

How much work must be done by a Carnot refrigerator to transfer \(3.5 \mathrm{~J}\) as heat (a) from a reservoir at \(7.0^{\circ} \mathrm{C}\) to one at \(35^{\circ} \mathrm{C}\), (b) from a reservoir at \(-73^{\circ} \mathrm{C}\) to one at \(35^{\circ} \mathrm{C}\), (c) from a reservoir at \(-173^{\circ} \mathrm{C}\) to one at \(35^{\circ} \mathrm{C}\), and (d) from a reservoir at \(-223^{\circ} \mathrm{C}\) to one at \(35^{\circ} \mathrm{C}\) ?

Figure 20-20 represents a Carnot engine that works between temperatures \(T_1=500 \mathrm{~K}\) and \(T_2=130 \mathrm{~K}\) and drives a Carnot refrigerator that works between temperatures \(T_3=360 \mathrm{~K}\) and \(T_4=200 \mathrm{~K}\). What is the ratio \(Q_3 / Q_1\) ?

An ice-cream maker is kept cold by a reverse Carnot engine that removes \(28.0 \mathrm{~kJ}\) as heat per cycle, with coefficient of performance 6.90. Per cycle, what are (a) the energy delivered as heat to the room and (b) the work done?

In an experiment, \(400 \mathrm{~g}\) of aluminum (with a specific heat of \(900 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\) ) at \(100^{\circ} \mathrm{C}\) is mixed with \(50.0 \mathrm{~g}\) of water at \(20.0^{\circ} \mathrm{C}\), with the mixture thermally isolated. (a) What is the equilibrium temperature? What are the entropy changes of (b) the aluminum, (c) the water, and (d) the aluminum- water system?

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.