A piston-cylinder device contains 3 kg of saturated water vapor at 200 kPa. Now heat is transferred to the system at constant pressure until the temperature reaches 200oC. Calculate the work done by the steam during this process.

work done = 117 kJ

Explanation:

given data

mass m = 3 kg

constant pressure P = 200 kPa

temperature T = 200°C

solution

we know that work done by steam is express as

work done = pressure × ΔVolume  1

and here ΔVolume  = final volume - initial volume

we use here steam table and get at pressure 200 kPa

final specific volume = 1.08052 m³/kg

and

initial specific volume = 0.885735 m³/kg

so here

ΔV  = 3 × (1.08052 - 0.885735)

ΔV  = 0.584 m³

so put value in equation 1 we get

work done by steam = 200 × 0.584

work done = 117 kJ

force is mass*acceleration.

we have a very small mass: 11.4*0.001 kg

if we want 0.035n of force, we need an acceleration:

0.035 = 11.4*0.001*a

a = 0.035/(11.4*0.001) m/s^2

the minimum amperage that the current in this wire would have to have so that its weight is completely supported by the magnetic force due to earth's field, assuming that no other forces except gravity act on it