A cylinder with a piston contains 0.8 kg of water at 100 degrees C. What is the change in internal energy of the water when it is converted to steam at 100 degrees C at a constant pressure of 1 atm? The density of water is 1000 kg/m^3 and that of steam is 0.6 kg/m^3. The latent heat of vaporization of water is 2.26E+06 J/kg.

1672.98 kJ (increase in internal energy)

Explanation:

Let's first find the energy received by the water in order for it to vaporize.

This is: Energy = mass of water * latent heat of vaporization

Energy = 0.8 * 2.26 * 10^6 = 1808 kJ

We also need to subtract from this the energy needed for the volume increase. This is calculated the following way:

Energy needed for volume change = P (Final Volume - Initial Volume)

Initial volume = 0.8/1000 = 0.0008 m^3

Final volume = 0.8 / 0.6 = 1.333 m^3

Pressure = 1 atm = 101325 Pa

Energy needed for volume change = 101325 (1.333 - 0.0008)

Energy needed for volume change = 135.02 kJ

Total Internal Energy Change = 1808 - 135.02 = 1672.98 kJ

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s