Question

For aluminum at \(0^{\circ} \mathrm{C}, \rho=2700 \mathrm{~kg} / \mathrm{m}^{3}, \beta=71.4 \times 10^{-8}\) \((\mathrm{K})^{-1}, \kappa=1.34 \times 10^{-13} \mathrm{~m}^{2} / \mathrm{N}\), and \(c_{p}=0.9211 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K} . \mathrm{De}-\) termine the percent error in \(c_{v}\) that would result if it were assumed that \(c_{p}=c_{v}\).

Short answer

The percent error in \(c_v\) assuming \(c_p = c_v\) is approximately 0.0417 \%.

Step-by-step solution

Identify the Given Values

List out the given values from the problem: \(\rho = 2700\, \text{kg/m}^3\), \(\beta = 71.4 \times 10^{-8}\, \text{K}^{-1}\), \(\kappa = 1.34 \times 10^{-13}\, \text{m}^2/\text{N}\), and \(c_p = 0.9211\, \text{kJ/kg} \cdot \text{K}\).

Convert Known Quantities to SI Units

Ensure all quantities are in SI units. \(c_p\) is already in \(\text{kJ/kg} \cdot \text{K}\), so convert it to \(\text{J/kg} \cdot \text{K}\): \[c_p = 0.9211 \times 1000 = 921.1 \text{J/kg} \cdot \text{K}\]

Calculate the Specific Heat at Constant Volume (\(c_v\))

Use the relationship between \(c_p\) and \(c_v\): \[ c_v = c_p - \left[ \frac{\beta^2 \cdot T \cdot V}{\kappa \cdot \rho} \right] \] Assume temperature \(T\) near \(0^{\circ} \text{C} \approx 0 + 273.15 = 273.15\, \text{K}\). The term \(\frac{\beta^2 \cdot T}{\kappa \cdot \rho}\) is: \[ \frac{(71.4 \times 10^{-8})^2 \cdot 273.15}{1.34 \times 10^{-13} \cdot 2700} \] Calculate the above term: \[ \text{Term} = \frac{(71.4^2) \times 10^{-16} \cdot 273.15}{1.34 \times 10^{-13} \cdot 2700} \approx \frac{5097.96 \times 10^{-16} \cdot 273.15}{3.618 \times 10^{-10}} \approx \frac{1391.5 \times 10^{-13}}{3.618 \times 10^{-10}} \approx 0.384 \text{J/kg} \cdot \text{K} \] Hence, \[ c_v \approx 921.1 - 0.384 = 920.716 \text{J/kg} \cdot \text{K} \]

Calculate the Percent Error

Find the percent error if assuming \(c_p = c_v\): \[ \text{Percent Error} = \left| \frac{c_p - c_v}{c_p} \right| \times 100 \% \] Substitute the values: \[ \text{Percent Error} = \left| \frac{921.1 - 920.716}{921.1} \right| \times 100 = \frac{0.384}{921.1} \times 100 \approx 0.0417 \% \]

Key concepts

headline of the respective core concept

Understanding specific heat capacities is crucial in thermodynamics.
Specific heat capacity \(c\) is the amount of energy needed to raise the temperature of 1 kg of a substance by 1 K.
There are two types: specific heat at constant pressure \(c_p\) and specific heat at constant volume \(c_v\).
For solids and liquids, we usually use \(c_p\), because their volumes often do not change significantly with temperature.
For gases, we often use \(c_v\), as their volumes can change drastically.
\(c_p\) is usually higher than \(c_v\) because at constant pressure, energy needs to do both heating and work to expand the substance.
For example, for aluminum:

• Density (\rho): 2700 kg/m³
• Thermal Expansion Coefficient (\beta): 71.4 × 10^(-8) K^(-1)
• Bulk Modulus (\kappa): 1.34 × 10^(-13) m²/N
• Specific Heat Capacity at Constant Pressure (c_p): 0.9211 kJ/kg·K

These values help us understand how aluminum responds to temperature changes.

headline of the respective core concept

The thermodynamic properties of materials describe how substances react to temperature, pressure, and volume changes.
These properties include density (\rho), thermal expansion coefficient (\beta), and bulk modulus (κ).

\beta measures how much a material's volume changes with temperature, and \(κ\) measures how much a material resists being compressed.
Understanding these properties helps in designing materials used in various applications, from household items to aerospace engineering.
When designing with aluminum, for instance, its high \(β\) means it will expand more than other metals when heated, which is crucial in designing joints and fixtures.

• Density (\rho): Indicates how much mass is stored in a unit volume of a material.
• Thermal Expansion Coefficient (\beta): Indicates how much a material's volume increases with temperature.
• Bulk Modulus (κ): Indicates a material’s resistance to uniform compression.

These properties work together to help us understand a material's overall behavior in different environments.

headline of the respective core concept

Percent error calculation is a measure to evaluate the accuracy of a value compared to a theoretical or true value.
It helps in understanding how far off an experimental value is from the accepted value.
The formula for calculating percent error is:
\text{Percent Error} = \left| \frac{\text{Experimental Value} - \text{Theoretical Value}}{\text{Theoretical Value}} \right| \times 100%\
In our example, if \( c_p = c_v \), you get: \text{Percent Error} = \left| \frac{921.1 - 920.716}{921.1} \right| \times 100 \approx 0.0417\text{%-} \.
This simple concept helps in interpreting and analyzing scientific experimental data.

headline of the respective core concept

The thermal expansion coefficient (\beta) is an important material property in thermodynamics, indicating how a material's volume changes with temperature.
For solids and liquids, \(β\) is tiny, but even small changes matter in engineering.
\(β\) is defined as:
\[ β = \frac{1}{V} \left( \frac{\partial V}{\partial T} \right)_p \] where \(V\) is the volume, and \(T\) is the temperature.
A higher \(β\) means the material expands more with temperature.
In our aluminum example, \(β = 71.4 \times 10^{-8} \text{ K}^{-1}\) means aluminum will slightly expand as temperature increases.
This is critical when designing objects that endure temperature fluctuations.