Homework Sourse085 kilograms of water at 200 kPa has internal energy of 245
0.85 kilograms of water at 200 kPa has internal energy of 2450kj. Determine temperature and specific volume.
0.85 kilograms of water at 200 kPa has internal energy of 2450kj. Determine temperature and specific volume.
Solution
internal energy = specific heat at given volue x change in temperature
2450= 4.186 x dt
dt = 585.20
t = 585.20 - 273
t = 312.20
pv=rt
200v=461.5 * 312.20
v = 720.4
