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

 Q = 3.33 108 J

Explanation:

This is an exercise in thermodynamics, specifically isobaric work

        W = V (- 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

when two balls of the different masses are dropped from equal height, they reach the ground at the same time .this is because they experience the same acceleration.

the acceleration is independent of mass.

thus when they are dropped they will always maintain the same velocity and travel the same distance (neglect air resistance of course) at the same time.