Question

g a 100 m3 container with 1/3 water with pressure of 100 MPa drops to 90 Mpa how much heat transfer is required to bring back to the initial condition

Answer 1

Q = 3.33 108 J

Step-by-step explanation:

This is an exercise in thermodynamics, specifically isobaric work

W = V (
`P_(f)`- P₀)

They tell us that we have ⅓ of the volume of the container filled with water

V = ⅓ 100

V = 33.3 m³

let's calculate

W = 33.3 (90-100) 10⁶

W = - 3.33 10⁸ J

To bring the system to its initial condition if we use the first law of thermodynamics

ΔE = Q + W

as we return to the initial condition the change of internal energy (ΔE) is zero

W = -Q

therefore the required heat is

Q = 3.33 108 J