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Debye's \(T^{3}\) Law. 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
(a) 36.73 J; (b) 0.734 J/mol·K; (c) 0.486 J/mol·K.

Step by step solution

01

Understand the Heat Capacity Formula

The molar heat capacity is given by the formula: \[ C = k \frac{T^3}{\Theta^3} \] where \( k = 1940 \mathrm{\ J/mol \cdot K} \) and \( \Theta = 281 \mathrm{\ K} \). This is used to calculate the amount of heat needed over a temperature range.
02

Integrate to Find Heat Required

The heat added, \( dQ \), is given by \( nC dT \). Integrate to find the total heat \( Q \) from the initial temperature \( T_1 \) to the final temperature \( T_2 \): \[ Q = \int_{T_1}^{T_2} n k \frac{T^3}{\Theta^3} dT \] For 1.50 moles, \( n = 1.50 \). Solve \[ Q = \frac{n k}{\Theta^3} \left[ \frac{T^4}{4} \right]_{T_1}^{T_2} \].
03

Calculate the Definite Integral

Substitute \( T_1 = 10 \) K and \( T_2 = 40 \) K: \[ Q = \frac{1.50 \times 1940}{281^3} \left( \frac{40^4}{4} - \frac{10^4}{4} \right) \]. Simplify and calculate the actual values to find \( Q \).
04

Evaluate Heat Required from Integration

Compute \( Q \) using the previous expression. It turns out to be approximately \( 36.73 \) J.
05

Determine Average Molar Heat Capacity

To find the average molar heat capacity, use: \[ \bar{C} = \frac{1}{T_2 - T_1} \int_{T_1}^{T_2} C \, dT \]. Simplify: \( \bar{C} = \frac{Q}{n(T_2-T_1)} \). Calculate \( \bar{C} \).
06

Calculate True Molar Heat Capacity at 40 K

Substitute \( T = 40 \) K into the heat capacity formula: \[ C = 1940 \times \frac{40^3}{281^3} \]. Calculate \( C \) at 40 K, which is approximately \( 0.486 \mathrm{\ J/mol \cdot K} \).

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Key Concepts

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

Molar Heat Capacity
Molar heat capacity is a fascinating property in physics and chemistry that indicates how much heat energy is needed to raise the temperature of one mole of a substance by one degree Kelvin (K). At low temperatures, Debye's T^3 Law becomes an excellent model for understanding how crystalline solids behave in terms of heat capacity. According to the law, the molar heat capacity is described by the equation \[C = k \frac{T^3}{\Theta^3}\]where the coefficient \(k\) is a constant related to the specific substance and \(\Theta\) represents the Debye temperature, a characteristic of the material system.In the context of Debye's T^3 Law:
  • The molar heat capacity increases with the cube of the temperature \(T\), meaning it's very low at low temperatures and increases as the temperature rises.
  • Rock salt, like many crystals, shows this behavior prominently, providing a typical example of Debye's theoretical predictions.
  • This law is particularly suitable for estimating how energy, in the form of heat, interacts with crystalline structures.
At higher temperatures, other factors overshadow Debye's Law, but in the low temperature domain explored here, it provides invaluable insights.
Heat Integration
Heat integration is a critical technique used to determine the total heat or thermal energy required for a specific process, often involving a temperature change. By using integration, we can cumulatively account for small changes in heat over a temperature range.In solving problems related to heat capacity using Debye's T^3 Law, such as the rock salt problem:
  • We begin by knowing that the exchanged heat \(dQ\) is equal to \(nC dT\), where \(n\) is the number of moles, \(C\) is the molar heat capacity, and \(dT\) is the small change in temperature.
  • This leads us to perform integration over the temperature path from \(T_1\) to \(T_2\):\[ Q = \int_{T_1}^{T_2} n \frac{kT^3}{\Theta^3} \, dT \]
  • Solving this equation provides us with the total thermal energy required, accounting for the nuanced ways temperature affects the heat capacity.
Heat integration shows the importance of calculus in providing precise calculations in thermodynamics challenges, offering a deeper understanding of how materials interact with energy experimentally and theoretically.
Low Temperature Physics
Low temperature physics explores the fascinating and often counterintuitive behavior of materials as they approach absolute zero. In this range, properties like heat capacity don't behave as they do at higher temperatures. This is where Debye's T^3 Law finds profound application. In the realm of low temperatures:
  • Heat capacity decreases significantly, meaning materials require less thermal energy to increase their temperature even by small amounts.
  • Crystalline structures, such as rock salt, exhibit unique heat capacity features that can be accurately modeled by Debye's approach.
  • Through empirical and theoretical frameworks like Debye’s, low temperature physics helps us understand the quantum effects that dominate at these temperatures, such as phonons (quantized sound waves) contributing to heat capacity.
Debye’s model becomes essential here, allowing scientists and engineers to not only predict but also manipulate material responses in ultra-cold environments, providing critical insights for applications in superconductivity, cryogenics, and quantum computing.

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Most popular questions from this chapter

Convert the following Kelvin temperatures to the Celsius and Fahrenheit scales: (a) the midday temperature at the surface of the moon \((400 \mathrm{K}) ;\) (b) the temperature at the tops of the clouds in the atmosphere of Saturn \((95 \mathrm{K}) ;(\mathrm{c})\) the temperature at the center of the sun \(\left(1.55 \times 10^{7} \mathrm{K}\right)\).

(a) If an area measured on the surface of a solid body is \(A_{0}\) at some initial temperature and then changes by \(\Delta A\) when the temperature changes by \(\Delta T,\) show that $$\Delta A=(2 \alpha) A_{0} \Delta T$$ where \(\alpha\) is the coefficient of linear expansion. (b) A circular sheet of aluminum is 55.0 \(\mathrm{cm}\) in diameter at \(15.0^{\circ} \mathrm{C}\) . By how much does the area of one side of the sheet change when the temperature increases to \(27.5^{\circ} \mathrm{C}\) ?

In a container of negligible mass, 0.0400 kg of steam at \(100^{\circ} \mathrm{C}\) and atmospheric pressure is added to 0.200 \(\mathrm{kg}\) of water at \(50.0^{\circ} \mathrm{C} .\) (a) If no heat is lost to the surroundings, what is the final temperature of the system? (b) At the final temperature, how many kilograms are there of steam and how many of liquid water?

Convert the following Celsius temperatures to Fahrenheit: (a) \(-62.8^{\circ} \mathrm{C},\) the lowest temperature ever recorded in North America (February \(3,1947,\) Snag, Yukon); (b) \(56.7^{\circ} \mathrm{C},\) the highest temperature ever recorded in the United States (July \(10,1913,\) Death Valley, California); \((\mathrm{c}) 31.1^{\circ} \mathrm{C},\) the world's highest average annual temperature (Lugh Ferrandi, Somalia).

A pot with a steel bottom 8.50 \(\mathrm{mm}\) thick rests on a hot stove. The area of the bottom of the pot is 0.150 \(\mathrm{m}^{2} .\) The water inside the pot is at \(100.0^{\circ} \mathrm{C},\) and 0.390 \(\mathrm{kg}\) are evaporated every 3.00 min. Find the temperature of the lower surface of the pot, which is in contact with the stove.

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