Given mass of Na is 23 g
Volume of water = [tex] 244 cm^{3} [/tex]
Mass of water = [tex] 244 cm^{3} * \frac{ 1 g}{cm^{3}} = 244 g [/tex]
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:
[tex] q = m C [/tex]ΔT
[tex] 71 kJ * \frac{1000 J}{1 kJ} = 267 g ( 4.18 \frac{J}{k.g} )(Tfinal - 293 K) [/tex]
[tex] 71,000 J = (1116.06 \frac{J}{K})(Tfinal - 293 K) [/tex]
Tfinal = 356.6 K
Therefore, the final temperature of water will be 356.6 K
