piston cylinder has 5kg water at 500kPa and 300c. Heated by 100V and 4A that operates for 500 secs. It's cooled to a saturated liquid. what is the boundary work and heat lost?

Answer: boundary work is 1.33kJ and heat lost is 200kJ

Explanation: mass of water m = 5kg

Molar mass of water mm = 18kg/mol

No of moles n = 5/18 = 0.28

Pressure P = 500x10^3

Temperature T = 300°c = 573K

R = 8.314pa. m3/mol/k a constant.

Using PV = nRT

V = nRT/P

V = (0.28x8.314x573) : (500x10^3)

V = 0.00266m3

Work on boundary w = PV

W = 500x10^3 x 0.00266 = 1330 J

= 1.33kJ

Heat lost on cooling is equal to electrical heat Q used to heat the water initially

Q = Ivt

I = current = 4A

v = voltage = 100

T = time = 500sec

Q = 4x100x500 = 200000 = 200kJ