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Molar heat capacity is an important concept in thermodynamics. It refers to the amount of heat needed to change the temperature of one mole of a substance by one degree Kelvin. This is written in the formula as \( C \), and it often varies with temperature for different substances. This variation is due to the changes in energies associated with the molecular configurations or states at different temperatures.

In our example, the molar heat capacity is expressed as:

• \( C = 29.5 \, \mathrm{J/mol \cdot K} + (8.20 \times 10^{-3} \, \mathrm{J/mol \cdot K^2}) \cdot T \)

This equation indicates that the molar heat capacity increases linearly with temperature \( T \). The first term is a constant, and the second term, containing \( T \), shows dependency on temperature. Understanding this variation is crucial for calculating the amount of heat needed for any thermodynamic process.

Temperature conversion is often necessary when dealing with heat capacity problems, since equations in thermodynamics typically use Kelvin. This conversion is straightforward, but essential, as the scales are different. To convert a temperature from Celsius to Kelvin, you simply add 273.15 to the Celsius temperature.

For instance, in the original problem, temperatures change from 27°C to 227°C, which in Kelvin is:

• Initial temperature: \( T_{i} = 27^{\circ} C + 273.15 = 300.15 \, K \)
• Final temperature: \( T_{f} = 227^{\circ} C + 273.15 = 500.15 \, K \)

This conversion ensures all calculations align with the international standard for thermodynamics, providing accuracy when integrating formulas and determining heat changes.

Integration is a powerful mathematical tool in thermodynamics, especially when dealing with equations where variables change. In the case of our exercise, integration helps calculate the total heat transfer over a temperature range.

• Start with the differential form \( dQ = nC dT \), where \( n \) is the number of moles.
• Integrate this over the limits of initial and final temperatures \( T_{i} \) and \( T_{f} \).

For our given formula, \( C = 29.5 + 8.20 \times 10^{-3} T \), integration over \( T_{i} \) to \( T_{f} \) is performed as follows:

• The integral becomes: \( \int_{T_{i}}^{T_{f}} (29.5 + 8.20 \times 10^{-3} T) \, dT = [29.5T + 4.10 \times 10^{-3} T^2]_{T_{i}}^{T_{f}} \)

This evaluation provides the heat transfer for one mole. Multiply by the number of moles, here 3, for the total heat required. Thus, integration simplifies how we account for varying factors like temperature over specific ranges, offering a precise solution.