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
