Heat Capacity

Overview

Engineering reference data for Heat Capacity in thermodynamics.

Key Formulas

First Law

$\Delta U = Q - W$

Energy is conserved — heat added minus work done.

Ideal Gas Law

$PV = nRT$

Relates pressure, volume, and temperature of an ideal gas.

Heat Transfer

$Q = mc\Delta T$

Sensible heat transfer.

Carnot Efficiency

$\eta = 1 - T_C/T_H$

Maximum efficiency between two temperatures.

Variables

Specific Heat for Gases

For vapors and gases, there are two distinct definitions of specific heat:

• Specific heat at constant pressure: $c_p = \left(\frac{\partial h}{\partial T}\right)_p$
• Specific heat at constant volume: $c_v = \left(\frac{\partial h}{\partial T}\right)_v$

Note: For solids and liquids, $c_p \approx c_v$ because these substances are nearly incompressible.

Gas Constant and Specific Heat Ratio

The individual gas constant $R$ relates the two specific heats:

$R = c_p - c_v$

The ratio of specific heats (also called the adiabatic index or isentropic expansion factor) is:

$k = \frac{c_p}{c_v}$

This ratio is critical in thermodynamic processes involving ideal gases, particularly in adiabatic expansions and compressions.

Molar Heat Capacity

Molar heat capacity ($C_p$) is the amount of heat required to raise the temperature of one mole of a substance by one degree at constant pressure.

Example: The molar heat capacity of iron is 25.10 J/(mol·K), meaning it takes 25.10 J of heat to raise 1 mol of iron by 1 K.

Converting Between Specific and Molar Heat Capacity

$c_p = \frac{C_p}{M} \qquad \text{and} \qquad C_p = c_p \cdot M$

where $M$ is the molar mass of the substance (g/mol).

Example — Methanol (CH₃OH):

Given: $C_p = 81.1$ J/(mol·K)

Molar mass: $M = 12.01 + 4(1.008) + 16.00 = 32.04$ g/mol

$c_p = \frac{81.1}{32.04} = 2.53 \text{ J/(g·K)}$

Unit Conversion Reference

Practical Examples

Heating Aluminum

Calculate the heat required to heat 2 kg of aluminum from 20 °C to 100 °C.

Given:

• Mass: $m = 2$ kg
• Specific heat: $c = 0.91$ kJ/(kg·°C)
• Temperature change: $\Delta T = 100 - 20 = 80$ °C

$Q = m \cdot c \cdot \Delta T = (2)(0.91)(80) = 145.6 \text{ kJ}$

Heating Water

Calculate the energy required to heat 1 liter of water from 0 °C to 100 °C.

Given:

• Volume: 1 L (mass = 1 kg, since density of water ≈ 1 kg/L)
• Specific heat: $c = 4.19$ kJ/(kg·°C)
• Temperature change: $\Delta T = 100 - 0 = 100$ °C

$Q = (1)(4.19)(100) = 419 \text{ kJ}$

Converting to kilowatt-hours:

$Q = 419 \text{ kJ} \times \frac{1}{3600} \text{ h/s} = 0.116 \text{ kWh}$

Engineering Insight: The high specific heat of water (4.19 kJ/kg·°C) compared to most metals makes it an excellent medium for thermal energy storage systems.

References

• Original Source