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First Law of Thermodynamics

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Thermodynamics is the branch of physical science that deal with the transformations of energy. The word thermodynamics is derived from the Greek, therm or heat, and dynamis or force.

The part of the universe that is at the centre of attention in thermodynamics is called the system. A system may be a block of iron, a beaker of water, an engine etc. The rest of the universe is called the surroundings. The surroundings are where we stand to make observations on the system and infer its properties.

Thermodynamics is based on a few basic postulates known as the laws of thermodynamics. The first law of thermodynamics point out the energy conservation. More details about the first law of thermodynamics is given in the following sections.
First Law of Thermodynamics

First Law of Thermodynamics Definition

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The statement of first law of thermodynamics is that 'energy can be neither created nor destroyed, although it can be transferred from one form to another'.
The expression for first law of thermodynamics is given by,
$\Delta $U = q + w
where,
U is the internal energy
q is the heat absorbed or released
w is the work done

Example of First Law of Thermodynamics

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Some of the examples of first law of thermodynamics are given below :
  • keep ice cubes in our hands until it melts : Consider you are the system and ice and rest of the universe is the surroundings. The heat will flow from our body to ice till it melts. So, the energy lost from our body and the energy gain to ice is same. Hence the first law is verified.
  • Vapourization of sweat from a human body : Here the system is sweat and surroundings is human body and other parts of universe. Heat flows from body to sweat in order to evaporate the sweat. So, in this case also energy is conserved.

First Law of Thermodynamics Problems

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Problems related to first law of thermodynamics are given in this section:

Solved Examples

Question 1: Consider a gas in a system has constant pressure and done a work of 500 J on to the system by releasing 60 J heat to the surrounding. Determine the internal energy of the system?

Solution:
 
Given that,
q = 60 J and w = 500 J

The equation for internal energy is,
$\Delta $U = q + w
$\Delta $U = 60 + 500 = 560 J

 

Question 2: Find out the internal energy of a system which has constant volume and the heat around the system is increased by 50 J?

Solution:
 
Given that,
q = 50 J

Since the gas is in constant volume, $\Delta $v=0
w = p $\Delta $v = 0

The equation for internal energy is,
$\Delta $U = q + w
$\Delta $U = q + 0
$\Delta $U = q = 50 J