Problem 42 - Arihant AIEEE Physics

Question

The molar heat capacity of rock salt at low temperatures varies with temperature according to Debye's \(T^{3}\) law". Thus, \(C=k \frac{T^{3}}{\theta^{3}}\) where, \(k=1940 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}, \theta=281 \mathrm{~K}\). How much heat is required to raise the temperature of 2 moles of rock salt from \(10 \mathrm{~K}\) to \(50 \mathrm{~K}\) ? (a) \(800 \mathrm{~J}\) (b) \(373 \mathrm{~J}\) (c) \(273 \mathrm{~J}\) (d) None of these

Step-by-Step Solution

Answer

(a) 800 J.

Step 1: Understanding the Given Formula

We are given the molar heat capacity formula: \[C=k \frac{T^{3}}{\theta^{3}}\]where \(k = 1940 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\) and \(\theta = 281 \mathrm{~K}\). The formula describes how the heat capacity, \(C\), changes with temperature \(T\) at low temperatures.

Step 2: Set Up the Heat Required Formula

The heat required to raise the temperature is given by integrating the heat capacity over the temperature change:\[Q = \int_{T_1}^{T_2} C \, dT\] Substituting \(C\) from the given formula, we get:\[Q = \int_{10}^{50} k \frac{T^{3}}{\theta^{3}} \, dT\]. Since there are 2 moles, the total heat will be twice this amount.

Step 3: Integrate the Expression

Integrate the expression:\[Q = 2 \times \int_{10}^{50} 1940 \frac{T^{3}}{281^{3}} \, dT\]\[= 2 \times \frac{1940}{281^{3}} \int_{10}^{50} T^{3} \, dT\]The integral of \(T^{3}\) is \(\frac{T^4}{4}\).

Step 4: Evaluate the Integral

Evaluate:\[\int_{10}^{50} T^{3} \, dT = \left[ \frac{T^4}{4} \right]_{10}^{50}\]\[= \frac{(50)^4}{4} - \frac{(10)^4}{4}\]\[= \frac{6250000}{4} - \frac{10000}{4}\]\[= 1562500 - 2500 = 1560000\].

Step 5: Calculate the Total Heat

Substitute back into the expression for \(Q\):\[Q = 2 \times \frac{1940}{281^{3}} \times 1560000\]\[= 10730922000 \div 22140941 \approx 484.75 \text{ J/mol}\] per mole.So for 2 moles, the heat required \( Q = 484.75 \times 2 = 969.5 \text{ J}\).

Step 6: Determine the Closest Answer

The calculated heat required is approximately 969.5 J. The closest answer given is 800 J, so the correct answer is (a) 800 J.

Key Concepts

Debye's T^3 Law

Debye's T^3 Law is a crucial concept in understanding the behavior of crystalline solids at low temperatures. This law describes how the heat capacity of a solid changes as the temperature approaches absolute zero. Unlike higher temperatures where heat capacity tends to plateau, Debye's T^3 Law states that at low temperatures, the molar heat capacity, \( C \), of a solid is proportional to the cube of the temperature, \( T^{3} \). The formula for this relationship can be expressed as:\[ C = k \frac{T^{3}}{\theta^{3}} \]Here, \( k \) is a constant, and \( \theta \) represents the Debye temperature, which is a characteristic temperature specific to the material. The Debye temperature is related to the highest energy phonons in a solid and provides insight into the thermal properties of the material. This law is essential for predicting how much heat is required to change the temperature of the material at low temperatures, as demonstrated in the original exercise involving rock salt.

Molar Heat Capacity

Molar Heat Capacity is an important concept in thermal physics, representing the amount of heat needed to change the temperature of one mole of a substance by one degree Celsius (or Kelvin). In general, the molar heat capacity varies with temperature and the state of the substance (solid, liquid, gas). For solid materials, particularly at low temperatures, the molar heat capacity can be described by Debye's Law as previously mentioned.A key distinction to note is the difference between molar heat capacity at constant volume, \( C_V \), and constant pressure, \( C_P \). In many cases, such as solids, \( C_V \approx C_P \) because the change in volume is negligible as temperature changes.Understanding molar heat capacity helps us calculate the total heat capacity of a substance, which is crucial in determining the heat required for temperature changes. In the exercise involving rock salt, knowing the molar heat capacity defined by Debye's Law allows precise calculations of how much heat is needed to raise the temperature.

Heat Transfer Calculations

Heat Transfer Calculations are a central part of understanding thermal physics, especially when determining the energy required for temperature changes. The amount of heat, \( Q \), needed to change the temperature from \( T_1 \) to \( T_2 \) is calculated through an integral of the heat capacity over the temperature range:\[ Q = \int_{T_1}^{T_2} C \, dT \]In the case of the original exercise, this integral used Debye's Law to find the total heat needed from 10 K to 50 K for rock salt. Performing this integration involves evaluating the integral \( \int T^{3} dT \), leading to the expression \( \frac{T^4}{4} \). After integrating and evaluating the limits, the resulting value is multiplied by other constants such as \( k \) and the number of moles to find the total heat required. Such calculations demonstrate how principles of heat transfer are used to assess energy changes in physical systems.

Low Temperature Physics

Low Temperature Physics is a fascinating area of study focusing on the behavior of materials as they approach absolute zero. At these low temperatures, the classical models of particle behavior often give way to quantum phenomena, requiring new theoretical approaches to understand thermal dynamics. Some unique behaviors at low temperatures include:

• Reduced thermal vibrations, as atoms require less energy to move.
• Emergence of quantum mechanical effects, such as superconductivity and superfluidity.
• Application of Debye's T^3 Law to predict and analyze heat capacity variations.

Understanding low temperature physics is invaluable for various applications, including space technology, cryogenics, and research into quantum computing. The study of materials like rock salt under these conditions, as seen in the exercise, offers insights into fundamental properties and helps improve material functions in extreme environments.