1. The first law of thermodynamics states that energy can be changed from one form to another, but cannot be created or destroyed. Heat and work are mutually convertible. 2. For a cyclic thermodynamic process, the net heat transfer equals the net work transfer. The first law must be satisfied at each step of the cycle. 3. Examples demonstrate using the first law to determine consistency of heat and work transfers for cyclic processes, and to calculate the magnitude and direction of heat transfer needed to satisfy the first law for a given work interaction and change in internal energy.

1. first law of
Thermodynamics
BY
Er. R.P. Ojha
1R.P. Sir

2. First law of thermodynamics
Statement 1
Energy can neither be created nor be destroyed, it can
be just changed from one form to another form
Statement 2
Heat and work are mutually convertible

3. First law for thermodynamic cycle
 Cyclic integral of heat is equal to
cyclic integral of work
Where
δQ = net amount of heat interaction
ΔW = net amount of work interaction

4. EXAMPLE
 A System undergoes a cycle composed of four
process.the heat transfer to each process are 400kJ,-
365kJ,-200kJ,25o kJ. The respective work transfer are
140kJ,0,-55kJ and 0. is data consistent with first law of
thermodynamics
δQ = 400+(-365)+(-200)+250 =85kJ
ΔW = 140+0+(-55)+0 = 85
Hence given data is consistent with first law of thermodynamics

5. Example
 The working fluid in a engineexecutes a cyclic process,
and two work interactions involved :10kJ to working
fluid and 30 kJ from the working fluid. The working
cycle also involves three heat transfer, two of which are
75kJ to working fluid and 40 kJ from working fluid.
Determine magnitude and direction of third
interaction
 -15kJ

6. First law of thermodynamics for process
 Amount of heat interaction
during a process is equal to sum
of amount of work interaction
and change in internal energy
 δQ = Δw + dU
 dU = change in internal energy

7. Example
 In a general compression process,2 kJ of mechanical
work is supplied to 5kg of working and 800 J of heat is
rejected to cooling jacket. Calcualte change in specific
internal energy
 δQ = Δw + dU
 240 J/kg