Question

A rotating paddle wheel is inserted in a closed pot of water. The stirring action of the paddle wheel heats the water. During the process, 30 kJ of heat is transferred to the water and 5 kJ is lost to the surrounding air. The paddle wheel work amounts to 500 J. Determine the final energy of the system if its initial energy is 10 kJ.

Answer 1

the final energy of the system is 35.5 kJ.

Step-by-step explanation:

Given;

initial energy of the system, E₁ = 10 kJ

heat transferred to the system, q₁ 30 kJ

Heat lost to the surrounding, q₂ = 5kJ

heat gained by the system, Q = q₁ - q₂ = 30 kJ - 5kJ = 25 kJ

work done on the system, W = 500 J = 0.5 kJ

Apply first law of thermodynamic,

ΔU = Q + W

where;

ΔU is change in internal energy

Q is the heat gained by the system

W is work done on the system

ΔU = 25kJ + 0.5 kJ

ΔU = 25.5 kJ

The final energy of the system is calculated as;

E₂ = E₁ + ΔU

E₂ = 10 kJ + 25.5 kJ

E₂ = 35.5 kJ.

Therefore, the final energy of the system is 35.5 kJ.