Given mass of Na is 23 g
Volume of water =
![244 cm^(3)](https://img.qammunity.org/2019/formulas/chemistry/high-school/o6l4h72le2c02ca491x6ahatnqeakuphx4.png)
Mass of water =
![244 cm^(3) * ( 1 g)/(cm^(3)) = 244 g](https://img.qammunity.org/2019/formulas/chemistry/high-school/w5efoxf0iejn47lcu4blfqh15qsgz4urts.png)
Total solution mass = 23 g + 244 g = 267 g
Specific heat capacity of water = 4.18 J/K.g
Equation relating mass, heat, specific heat capacity and temperature change is:
ΔT
![71 kJ * (1000 J)/(1 kJ) = 267 g ( 4.18 (J)/(k.g) )(Tfinal - 293 K)](https://img.qammunity.org/2019/formulas/chemistry/high-school/xl0h8jdzlocmcnxfjq91nwawgi0qs3ipqg.png)
![71,000 J = (1116.06 (J)/(K))(Tfinal - 293 K)](https://img.qammunity.org/2019/formulas/chemistry/high-school/drtmq37e1jmgeayo47jt2qbg6o3jnpncft.png)
Tfinal = 356.6 K
Therefore, the final temperature of water will be 356.6 K