# The cyclic integral of (δQ − δW) for a process is

### Right Answer is:

Zero

#### SOLUTION

It was established by an English Scientist James Prescott Joule (1819-83) in 1843, that heat and mechanical energies are mutually convertible. He established through experiments that whenever a certain amount of work is done, a definite quantity of heat is produced. In other words, there is a numerical relation between the unit of work and a unit of heat. This relation is denoted by J (named after Joule) and it is known as Joule’s equivalent of the mechanical equivalent of heat.

Since the unit of work is joule or kilojoule and the unit of heat is also joule or kilojoule, therefore we can convert straightway heat units into mechanical units and vice versa. The relationship between heat and work is the basis of the First Law of Thermodynamics. It is one of the most fundamental laws of energy and is of special importance to engineers. It is also called the law of conservation of energy.

**First Law of Thermodynamics**

According to this law, when a closed or non-flow system undergoes a thermodynamic cycle, the net heat transfer is equal to the network transfer. Mathematically,

∮Q = ∮W——-(1)

i.e. the cyclic integral of heat transfer is equal to the cyclic integral of work transfer. It may be noted that the symbol ∮ stands for cyclic integral (i.e. integral around a complete cycle), δQ & δW are small elements of heat and work transfers respectively, and have the same units. The equation (i) may be written as

**∮(δQ − δW) = 0**

It follows from this equation that “if a system is taken through a cycle of processes so that it returns to the same state or condition from which it started, then the sum of heat and work will be zero.” It may be noted that δQ & δW are both path functions, but their difference ie. (δQ − δW) is a point function as the cyclic integral of (δQ − δW) is zero, i.e. ∮(δQ − δW) = 0.