This implies that (Î´W-Î´Q) must be a point function or property of the system. This property is termed as internal energy, U. Mathematically, internal energy can be written as:

**First law of thermodynamics for a closed system:**_{1}Q

_{2 }amount of heat is transferred across its boundary and

_{1}W

_{2 }is the amount of work done by the system and the system is allowed to come to an equilibrium state 2. Then integration of above Equation yields,

*m*is the mass of the system and

*u*denotes the specific internal energy of the system then,

_{1}q

_{2}and

_{1}w

_{2 }are heat transfer and work done per unit mass of the system.

**Flow Work:**

In an open system some
matter, usually fluid enters and leaves the system. It requires flow work for the fluid to enter the system against the system pressure and at the same
time flow work is required to expel the fluid from the system. It can be shown that the specific
flow work is given by the product of pressure, *p *and specific volume, *v*, i.e.,

Flow
work = *pv*.

**Enthalpy:***pv’*with internal energy ‘

*u’*as both of them increase the energy of the system. The sum of specific internal energy and specific flow work is an intensive property of the system and is called specific enthalpy, h. Thus specific enthalpy, h is given by:

**First law of thermodynamics for open system:**

__First law for open system in steady state:__

In steady state process, the time rate of change of all the quantities is zero, and mass is also conserved. As a result, the mass and total energy of the system do not change with time, hence, (dE/dt) is zero and from conservation of mass, m1 = m2 = m. Then the first law becomes:

Where, q and w are specific heat and work transfer rates.

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