According to the first law of thermodynamics, while doing external work, some amount of heat is given to the system. Then,
Amount of the heat absorbed = Sum of the increase in internal energy of the system
It is due to an increase in temperature, as well as the external work done by the system during the process of expansion.
∆Q Heat supply
In simple words, the first law of thermodynamics is defined as
Heat supplied to system = sum of the changes in internal energy + work done by the system
The first law of thermodynamics equation is represented by
ΔU =ΔQ – ΔW
ΔU represents a change in internal energy of the thermodynamic system
ΔQ represents heat given to the system
ΔW represents work done
If we differentiate the above equation, we get dU=dQ-dW
U (internal energy)
At molecular level
Energy = Sum of K.E (kinetic energy) and potential energy
Here are 3 types of K.E and three types of potential energy are given below
U=translational motion(kinetic energy) + Vibrational motion(Kinetic energy) + Rotational motion(Kinetic energy) + Force of attraction of electrons and nucleus(Potential energy) + binding energy of molecules
U (internal energy) depends on the final and initial state of the system. It is independent of path.
dU = nCvΔT
where,
Q (Heat)
If the heat is added to the system, it is positive, and if the heat is released from the system, it is negative
W (Work)
Relation: Δ U = Q − W .
To understand much better, here is an example.
Consider a heat engine that converts thermal energy to mechanical energy and vice versa. Heat engines have different types of relationships with heat, pressure, and the volume of the working fluid, which is normally a gas. Here, energy is transferred during phase changes from liquid to gas, or vice versa. In such cases, energy is not created.
FOR EXAMPLE:
Solution:
ΔU = Q – W
ΔU = 150J – (- 400J)
ΔU = 550J
ΔU is the internal energy relationship that revolves between the system and surroundings. If the surroundings lose some energy, the system gains the lost energy. Moreover, the surrounding area will also lose some heat to carry out work in the system.
Here are a few applications of the first law of thermodynamics in real life :
If the door of the refrigerators is kept open for a long period, then the entire room’s temperature will rise.
Here are some other applications
The ideal gas temperature remains the same in the case of an isothermal process (dU = 0) so that
dQ = dU + dW ⇒ dQ = dW.
Internal energy increases when solid melts to liquid, where m is the mass of liquid and L is the latent heat of the solid. In the system, the heat absorbed is :
dQ = mL
where the same expansion is, change in volume = 0 so,
⇒ dW = PΔV = 0
so, dQ = dU + dW ⇒ dU = mL
Internal energy increases during the melting process
∆Q (+ve)
∆W (+ve)
∆Q (-ve) ∆W (-ve)
| Work done | Heat | Sign |
| Work done By the system | Heat gained by the system |
∆Q= +ve ∆W= +ve |
| Work done by the system | Heat lost by the system |
∆Q= -ve ∆W= +ve |
| WorK done on the system | Heat gained by the system |
∆Q=+ve ∆W=-ve |
| Work done on the system | Heat lost by the system |
∆Q=-ve ∆W=-ve |