**Internal Energy Formula**

Internal Energy Formula is the heat energy stocked in gas. If a certain amount of heat is applied to gas, the result is that the temperature of the gas may increase or else the volume of gas might increase. In this manner doing some work externally or volume and temperature may both intensify, but it will be made definite by the situations under which the gas is provided heat.

When the structure is left in a situation which differs from its condition before the process, after the process, it can be articulated as

ΔΔU = U2 – U1 = Q – wt/J

Or else

Q = (U2 – U1) + W

U1 is termed as the internal energy of the system at the commencement of the procedure, and the internal energy at the culmination of the procedure is U2.

**Solved Problems**

**Problem 1**: Calculate the change in internal energy of the working fluid stating whether it is a gain or loss when in an internal combustion engine, the heat rejected to the cooling water during the compression stroke is 50 KJ/kg and the work input is 100 KJ/kg.

**Answer:**

Heat given to the cooling water

Work input W = – 100 KJ/kg

Q = – 50 KJ/kg

Using the formula,

Q = (U2 – U1) + W

– 50 = (U2 – U1) – 100

U2 – U1 = – 50 + 100

U2 – U1 = 50 KJ/kg

Therefore, the increases in internal energy is = 50 KJ/kg

**Problem 2**: The compressed air in an air motor cylinder has an internal energy of 450 KJ/kg at the commencement of the expansion and an internal energy of 220 KJ/kg afterwards the expansion. If the effort completed by the air throughout the expansion is 120 KJ/kg, compute the heat flow back and forth from the cylinder.

**Answer:**

Internal energy at commencement of the expansion

U1 = 450 KJ/kg

Internal energy afterwards the expansion

U2 = 220 KJ/kg

Work done by the air for the duration of the expansion

W = 120 KJ/kg

Calculate the Heat flow Q by means of the equation = (U2 – U1) + W

Q = (220 – 450) + 120

Q = – 230 + 120

= – 110 KJ/kg

Thus, the heat rejected by the air is 110 KJ/kg

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