/*! 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 85 At very low temperatures the mol... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

At very low temperatures the molar heat capacity of rock salt varies with temperature according to Debye's \(T^{3}\) law: $$ C=k \frac{T^{3}}{\theta^{3}} $$ where \(k=1940 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K}\) and \(\theta=281 \mathrm{~K}\). (a) How much heat is required to raise the temperature of \(1.50 \mathrm{~mol}\) of rock salt from \(10.0 \mathrm{~K}\) to \(40.0 \mathrm{~K} ?\) (Hint: Use Eq. (17.18) in the form \(d Q=n C d T\) and integrate.) (b) What is the average molar heat capacity in this range? (c) What is the true molar heat capacity at \(40.0 \mathrm{~K} ?\)

Short Answer

Expert verified
The amount of heat required to raise the temperature of 1.50 mol of rock salt from 10.0 K to 40.0 K can be calculated using the first integral in Step 2. The average heat capacity can be calculated using the formula in Step 3. The specific heat capacity at 40.0 K can be calculated using the formula given in Step 4.

Step by step solution

01

Understand Basic Concepts

Debye's Law models the variation of molar heat capacity, \(C\), with temperature, \(T\). \[C=k \frac{T^{3}}{\theta^{3}}\], where \(k=1940 \mathrm{~J} / \mathrm{mol} . \mathrm{K}\) and \(\theta=281 \mathrm{~K}\). The equation \(d Q=n C d T\) shows heat needed for temperature change of material with molar heat capacity C.
02

Calculate Total Heat Required to Go from 10.0K to 40.0K

Given that heat=\(d Q=n C d T\) and \(C=k \frac{T^{3}}{\theta^{3}}\), substitute C into the heat equation to get \(d Q=n k \frac{T^{3}}{\theta^{3}} d T\). To calculate heat, integrate from \(10.0K\) to \(40.0K\): \[Q = \int_{10}^{40} n k \frac{T^{3}}{\theta^{3}} dT \]
03

Calculate Average Heat Capacity

The average molar heat capacity is the total heat divided by the total number of moles and the total temperature change. This is calculated as: \[ \frac{Q}{n \times (T_{f}-T_{i})}\] where \(T_{f} = 40.0 K\) and \(T_{i} = 10.0 K\)
04

Find Heat Capacity at 40.0K

Substitute \(T=40.0K\) into Debye's law to find the heat capacity at that specific temperature: \[ C = k\frac{T^{3}}{\theta^{3}} = k\frac{(40.0K)^{3}}{(281K)^{3}} \]

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.

Debye's Law
Debye's Law is a principle in solid-state physics that describes how the molar heat capacity of a substance changes with temperature, especially at very low temperatures. At these temperatures, the heat capacity doesn't increase linearly with temperature, as it might do at higher temperatures, but rather proportionally to the cube of the temperature. This is expressed mathematically as:

\[ C=k \frac{T^3}{\theta^3} \]
Where:\
  • \( C \) is the molar heat capacity at a constant volume.
  • \( k \) is a constant specific to the material.
  • \( T \) is the absolute temperature.
  • \( \theta \) (theta) is the Debye temperature, a characteristic temperature for the substance.

Debye's Law predicts that at very low temperatures, the heat capacity of crystalline solids will drop off significantly as the temperature approaches zero. This model is particularly useful for understanding the behavior of materials at temperatures near absolute zero.
Thermodynamics
Thermodynamics is the branch of physics that deals with the relationships between heat and other forms of energy. In the context of our discussion, thermodynamics helps us understand how heat transfer results in changes in temperature and physical state of a substance.

The first law of thermodynamics, also known as the law of energy conservation, states that energy cannot be created or destroyed in an isolated system. This principle underlies the calculations for heat required to raise the temperature of a substance. The specific heat capacity of a material is a measure of how much heat energy is needed to raise the temperature of a certain amount of the substance by a certain amount. When we apply the first law of thermodynamics to the problem of calculating heat required to change a substance's temperature, we talk about integrating the heat capacity over a temperature range, which gives us the total heat (or energy) change.
Heat Capacity Integration
Heat capacity integration is a mathematical method used to calculate the total heat required to change a substance's temperature over a range of temperatures. The molar heat capacity can vary with temperature, so simply multiplying it by the temperature change and the amount of substance doesn't always give the correct answer.

Instead, we integrate the molar heat capacity function over the temperature range of interest:

\[Q = \int_{T_i}^{T_f} nC(T) dT \]
Where:\
  • \( Q \) is the total heat required.
  • \( n \) is the number of moles of the substance.
  • \( C(T) \) is the temperature-dependent molar heat capacity.
  • \( T_i \) and \( T_f \) are the initial and final temperatures.

This integration accounts for how the molar heat capacity may change as the temperature changes, giving a more accurate measure of the total heat required for the temperature change.
Low Temperature Heat Capacity
Low temperature heat capacity refers to the behavior of a material's heat capacity at temperatures close to absolute zero. As a material's temperature approaches zero, heat capacity tends to also approach zero, which is explained by Debye's Law and the third law of thermodynamics.

The third law of thermodynamics states that as the temperature of a perfect crystal approaches absolute zero, its entropy, which is a measure of disorder, approaches a constant minimum. This has implications for the heat capacity, which is linked to changes in entropy with temperature. At very low temperatures, the quantization of energy levels becomes significant, and the heat capacity becomes a function of those discrete energy levels, rather than a continuous function of temperature. This is why the heat capacity drops off as described by Debye's model.

The study of low-temperature heat capacity is not just academic; it has practical applications in fields such as cryogenics and materials science, where understanding how substances behave at extreme cold is crucial.

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

In an effort to stay awake for an all-night study session, a student makes a cup of coffee by first placing a \(200 \mathrm{~W}\) electric immersion heater in \(0.320 \mathrm{~kg}\) of water. (a) How much heat must be added to the water to raise its temperature from \(20.0^{\circ} \mathrm{C}\) to \(80.0^{\circ} \mathrm{C} ?\) (b) How much time is required? Assume that all of the heater's power goes into heating the water.

Consider a poor lost soul walking at \(5 \mathrm{~km} / \mathrm{h}\) on a hot day in the desert, wearing only a bathing suit. This person's skin temperature tends to rise due to four mechanisms: (i) energy is generated by metabolic reactions in the body at a rate of \(280 \mathrm{~W},\) and almost all of this energy is converted to heat that flows to the skin; (ii) heat is delivered to the skin by convection from the outside air at a rate equal to \(k^{\prime} A_{\mathrm{skin}}\left(T_{\mathrm{air}}-T_{\mathrm{skin}}\right),\) where \(k^{\prime}\) is \(54 \mathrm{~J} / \mathrm{h} \cdot \mathrm{C}^{\circ} \cdot \mathrm{m}^{2},\) the exposed skin area \(A_{\text {skin }}\) is \(1.5 \mathrm{~m}^{2},\) the air temperature \(T_{\text {air }}\) is \(47^{\circ} \mathrm{C},\) and the skin temperature \(T_{\text {skin }}\) is \(36^{\circ} \mathrm{C} ;\) (iii) the skin absorbs radiant energy from the sun at a rate of \(1400 \mathrm{~W} / \mathrm{m}^{2} ;\) (iv) the skin absorbs radiant energy from the environment, which has temperature \(47^{\circ} \mathrm{C}\). (a) Calculate the net rate (in watts) at which the person's skin is heated by all four of these mechanisms. Assume that the emissivity of the skin is \(e=1\) and that the skin temperature is initially \(36^{\circ} \mathrm{C}\). Which mechanism is the most important? (b) At what rate (inL/h) must perspiration evaporate from this person's skin to maintain a constant skin temperature? (The heat of vaporization of water at \(36^{\circ} \mathrm{C}\) is \(2.42 \times 10^{6} \mathrm{~J} / \mathrm{kg} .\) ) (c) Suppose the person is protected by light-colored clothing \((e \approx 0)\) and only \(0.45 \mathrm{~m}^{2}\) of skin is exposed. What rate of perspiration is required now? Discuss the usefulness of the traditional clothing worn by desert peoples.

You are given a sample of metal and asked to determine its specific heat. You weigh the sample and find that its weight is \(28.4 \mathrm{~N}\). You carefully add \(1.25 \times 10^{4} \mathrm{~J}\) of heat energy to the sample and find that its temperature rises \(18.0 \mathrm{C}^{\circ} .\) What is the sample's specific heat?

\(17.76 \bullet\) A surveyor's \(30.0 \mathrm{~m}\) steel tape is correct at \(20.0^{\circ} \mathrm{C}\). The distance between two points, as measured by this tape on a day when its temperature is \(5.00^{\circ} \mathrm{C},\) is \(25.970 \mathrm{~m} .\) What is the true distance between the points?

A U.S. penny has a diameter of \(1.9000 \mathrm{~cm}\) at \(20.0^{\circ} \mathrm{C}\). The coin is made of a metal alloy (mostly zinc) for which the coefficient of linear expansion is \(2.6 \times 10^{-5} \mathrm{~K}^{-1}\). What would its diameter be on a hot day in Death Valley \(\left(48.0^{\circ} \mathrm{C}\right) ?\) On a cold night in the mountains of Greenland \(\left(-53^{\circ} \mathrm{C}\right) ?\)

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.