How did we get H'₂? I don't know how they interpolated, please help me with this.

Example 7.8 Saturated-vapor steam at 100kPa(tsat=99.63C) is compressed adiabatically to 300kPa. If the compressor efficiency is 0.75, what is the work required and what are the properties of the discharge stream? For saturated steam at 100kPa, S1=7.3598kJkg1K1H1=2675.4kJkg1 For isentropic compression to 300kPa,S2=S1=7.3598kJkg1K1. Interpolation in the tables for superheated steam at 300kPa shows that steam with this entropy has the enthalpy: H2=2888.8kJkg1. ? By Eq. (7.17), (H)=(H)S=0.75213.4=284.5kJkg1 and H2=H1+H=2675.4+284.5=2959.9kJkg1 For superheated steam with this enthalpy, interpolation yields: T2=246.1CS2=7.5019kJkg1K1 Moreover, by Eq. (7.14), the work required is: Ws=H=284.5kJkg1