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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

Sagot :

Answer:

 Q = 3.33 108 J

Explanation:

This is an exercise in thermodynamics, specifically isobaric work

        W = V ([tex]P_{f}[/tex]- 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

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