For the steady flow process, the first law is written like
DH + Du2/2 + gDz = Q + Ws
since there is no shaft work, Ws = 0
and flow is horizontal, Dz = 0
Therefore,
DH + Du2/2 = Q
substituting for the quantities,
(2726.5 - 334.9) x 1000 + (200^2 - 3^2)/2 = Q (in terms of J/kg)
Q = 2411.1 kJ/kg
Heat transferred through the coil per unit mass of water = 2411.1 kJ