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1.5 kg of water is in saturated liquid state at 2 bar (vf = 0.001061 m3/kg, uf = 504.0 kJ/kg, hf = 505 kJ/kg). Heat is added in a constant pressure process till the temperature of water reaches 400°C (v = 1.5493 m3/kg, u = 2967.0 k J/kg, h = 3277.0 kJ/kg). The heat added (in kJ) in the process is______.
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- 4158
- 4258
- 5158
- 4168
Correct Option: A
Given, m = 1.5 kg
h1 = hf = 505 kJ/kg h2 = 3277.0
From Ist Law,
dQ = du + pdv = dh – vdp
dQ = dh (as vdp = 0)
Qadd = dQ = m (h2 – h1) = (3277.0 – 505) × 1.5
Qaddcd = 4158 kJ.
