## Enthalpy in Extensive Units

**H = U + pV**

**Enthalpy** is an extensive quantity, it depends on the size of the system, or on the amount of substance it contains. The SI unit of enthalpy is the joule (J). It is the energy contained within the system, excluding the kinetic energy of motion of the system as a whole and the potential energy of the system as a whole due to external force fields. It is the thermodynamic quantity equivalent to the **total heat content** of a system.

On the other hand, energy can be stored in the chemical bonds between the atoms that make up the molecules. This energy storage on the atomic level includes energy associated with electron orbital states, nuclear spin, and binding forces in the nucleus.

**Enthalpy** is represented by the symbol **H**, and the change in enthalpy in a process is **H _{2} – H_{1}**.

There are expressions in terms of more familiar variables such as temperature and pressure:

**dH = C _{p}dT + V(1-αT)dp**

Where **C _{p}** is the

**heat capacity at constant pressure**and

**is the coefficient of (cubic) thermal expansion. For ideal gas αT = 1 and therefore:**

*α** dH = C _{p}dT*

## Example: Frictionless Piston – Heat – Enthalpy

A frictionless piston is used to provide a constant pressure of **500 kPa** in a cylinder containing steam (superheated steam) of a volume of **2 m ^{3 }** at

**500 K**. Calculate the final temperature, if

**3000 kJ**of

**heat**is added.

**Solution:**

Using steam tables we know, that the **specific enthalpy** of such steam (500 kPa; 500 K) is about** 2912 kJ/kg**. Since at this condition the steam has density of 2.2 kg/m^{3}, then we know there is about **4.4 kg of steam** in the piston at enthalpy of 2912 kJ/kg x 4.4 kg =** 12812 kJ**.

When we use simply **Q = H _{2} − H_{1}**, then the resulting enthalpy of steam will be:

H_{2} = H_{1} + Q = **15812 kJ**

From **steam tables**, such superheated steam (15812/4.4 = 3593 kJ/kg) will have a temperature of **828 K (555°C)**. Since at this enthalpy the steam have density of 1.31 kg/m^{3}, it is obvious that it has expanded by about 2.2/1.31 = 1.67 (+67%). Therefore the resulting volume is 2 m^{3} x 1.67 = 3.34 m^{3} and ∆V = 3.34 m^{3} – 2 m^{3} = 1.34 m^{3}.

The **p∆V ** part of enthalpy, i.e. the work done is:

** W = p∆V = 500 000 Pa x 1.34 m ^{3} = 670 kJ**

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