Question

The constant-pressure specific heat of air at \(25^{\circ} \mathrm{C}\) is \(1.005 \mathrm{kJ} / \mathrm{kg} \cdot^{\circ} \mathrm{C} .\) Express this value in \(\mathrm{kJ} / \mathrm{kg} \cdot \mathrm{K}, \mathrm{J} / \mathrm{g} \cdot^{\circ} \mathrm{C}, \mathrm{kcal} /\) \(\mathrm{kg} \cdot^{\circ} \mathrm{C},\) and \(\mathrm{Btu} / \mathrm{lbm} \cdot^{\circ} \mathrm{F}\).

Short answer

Question: Convert the constant-pressure specific heat of air at 25°C (1.005 kJ/kg·°C) to the following units: - kJ/kg·K - J/g·°C - kcal/kg·°C - Btu/lbm·°F Answer: - 1.005 kJ/kg·K - 1.005 J/g·°C - 0.2402 kcal/kg·°C - 0.2404 Btu/lbm·°F

Step-by-step solution

Convert to \(\mathrm{kJ} / \mathrm{kg} \cdot \mathrm{K}\)

Since \(1^{\circ} \mathrm{C}\) change is equivalent to \(1 \mathrm{K}\) change, the given value is already in \(\mathrm{kJ} / \mathrm{kg} \cdot \mathrm{K}\). So, the value is \(1.005 \mathrm{kJ} / \mathrm{kg} \cdot \mathrm{K}\).

Convert to \(\mathrm{J} / \mathrm{g} \cdot^{\circ} \mathrm{C}\)

To convert \(\mathrm{kJ}\) to \(\mathrm{J}\), we multiply by \(1000\). To convert \(\mathrm{kg}\) to \(\mathrm{g}\), we multiply by \(1000\) as well. So, the value is \(1.005 \times 1000 \mathrm{J} / (1000 \mathrm{g} \cdot^{\circ} \mathrm{C}) = 1.005 \mathrm{J} / \mathrm{g} \cdot^{\circ} \mathrm{C}\).

Convert to \(\mathrm{kcal} / \mathrm{kg} \cdot^{\circ} \mathrm{C}\)

We can use the conversion factor \(1 \mathrm{kcal} = 4184 \mathrm{J}\) to convert energy from \(\mathrm{J}\) to \(\mathrm{kcal}\). So, the value is \(1.005 \mathrm{kJ} / \mathrm{kg} \cdot^{\circ} \mathrm{C} \times (\frac{1 \mathrm{kcal}}{4184 \mathrm{J}}) = 0.2402 \mathrm{kcal} / \mathrm{kg} \cdot^{\circ} \mathrm{C}\) (rounded to four decimal places).

Convert to \(\mathrm{Btu} / \mathrm{lbm} \cdot^{\circ} \mathrm{F}\)

We need two conversion factors for this step: \(1 \mathrm{Btu} = 1055.06 \mathrm{J}\) (for energy) and \(1 \mathrm{lbm} = 0.453592 \mathrm{kg}\) (for mass). We also need the relationship between Fahrenheit and Celsius, which is \(\Delta T_{\mathrm{F}} = \frac{9}{5} \Delta T_{\mathrm{C}}\) (for a change in temperature difference). So, we convert the value as follows: \(1.005 \mathrm{kJ} / \mathrm{kg} \cdot^{\circ} \mathrm{C} \times (\frac{1 \mathrm{Btu}}{1055.06 \mathrm{J}}) \times (\frac{1 \mathrm{lbm}}{0.453592 \mathrm{kg}}) \times (\frac{5^{\circ}\mathrm{F}}{9^{\circ}\mathrm{C}}) = 0.2404 \mathrm{Btu} / \mathrm{lbm} \cdot^{\circ} \mathrm{F}\) (rounded to four decimal places). The constant-pressure specific heat of air at \(25^{\circ} \mathrm{C}\) can now be expressed in all the desired units: - \(1.005 \mathrm{kJ} / \mathrm{kg} \cdot \mathrm{K}\) - \(1.005 \mathrm{J} / \mathrm{g} \cdot^{\circ} \mathrm{C}\) - \(0.2402 \mathrm{kcal} / \mathrm{kg} \cdot^{\circ} \mathrm{C}\) - \(0.2404 \mathrm{Btu} / \mathrm{lbm} \cdot^{\circ} \mathrm{F}\)

Key concepts

Constant-Pressure Specific Heat

Understanding the constant-pressure specific heat of a substance is vital in many engineering and physics applications. This value indicates the amount of energy required to raise the temperature of a unit mass of a substance by one degree at constant pressure. For example, the constant-pressure specific heat of air at a certain temperature, such as \(25^{\circ} \mathrm{C}\), defines how much energy in kilojoules is needed to increase the temperature of one kilogram of air by one degree Celsius or one kelvin under constant pressure.

This concept is crucial when dealing with energy transfer in systems where the volume can change, such as in the study of gases in thermodynamics. When a student encounters a problem involving specific heat, it's important to pay attention to the pressure condition—constant-volume specific heat values will differ from constant-pressure specific heat values for the same substance.

Temperature Unit Conversion

Temperature units like Celsius (\( ^{\circ} \)C), Kelvin (K), and Fahrenheit (\( ^{\circ} \)F) can often be a source of confusion. Converting between these units is a common task in science and engineering. The Kelvin scale is an absolute temperature scale used primarily in scientific settings, where 0 K is absolute zero. Celsius, used commonly in many countries for day-to-day temperature measurements, has a direct conversion to Kelvin: \( K = C + 273.15 \). Fahrenheit, more commonly used in the United States for everyday temperatures, has a more complex conversion to Celsius: \( C = (F - 32) \times \frac{5}{9} \).

Key Conversions:

• From Fahrenheit to Celsius: \( C = (F - 32) \times \frac{5}{9} \)
• From Celsius to Fahrenheit: \( F = C \times \frac{9}{5} + 32 \)
• From Celsius to Kelvin: \( K = C + 273.15 \)
• From Kelvin to Celsius: \( C = K - 273.15 \)

These conversions are essential when dealing with scientific data or working in an environment that uses different units of measurement.

Energy Unit Conversion

Energy units such as joules (J), kilojoules (kJ), calories (cal), and British Thermal Units (Btu) are often interchangeably used in physics and engineering. Understanding how to convert between these units enables students to solve problems across various disciplines and understand energy-related concepts. The joule is the standard unit of energy in the International System of Units (SI), while the calorie is often used in chemistry and biology, and the Btu is commonly used in heating and cooling systems.

Key Conversions:

• From joules to kilojoules: \( 1 kJ = 1000 J \)
• From kilojoules to joules: \( 1 J = 0.001 kJ \)
• From joules to calories: \( 1 cal = 4.184 J \)
• From calories to joules: \( 1 J = 0.239 cal \)
• From joules to Btu: \( 1 Btu = 1055.06 J \)
• From Btu to joules: \( 1 J = 0.000947817 Btu \)

When converting energy units, it is also essential to consider the mass and temperature units in use, as these can affect the conversion factor. Being proficient in these conversions is critical when analyzing energy transfer, studying thermodynamics, or working with heating and ventilation systems.