Answer: 11,000 J
Explanation:
In an isothermal process,

(1)
Note that, the energy used in heat transfer is not available for work. So, the amount of energy unavailable for work is equal to the energy used in heat transfer.
To obtain the amount of energy in heat transfer, we multiply both sides of equation (1) by the denominator of the right side of (1) so that
amount of energy in heat transfer = (entropy increase)(temperature)
= (25 J/K)(440 K)
= 11,000 J
Since the amount of energy unavailable for work is equal to the amount of energy in the heat transfer, therefore the amount of energy unavailable for work is 11,000 J.
That would be the presence of a magnet anywhere near the compass.
To solve this problem it is necessary to apply the continuity equations in the fluid and the kinematic equation for the description of the displacement, velocity and acceleration.
By definition the movement of the Fluid under the terms of Speed, acceleration and displacement is,

Where,
Velocity in each state
g= Gravity
h = Height
Our values are given as,



Replacing at the kinetic equation to find
we have,



Applying the concepts of continuity,

We need to find A_2 then,

So the cross sectional area of the water stream at a point 0.11 m below the faucet is



Therefore the cross-sectional area of the water stream at a point 0.11 m below the faucet is 