Question

Question #24eab

Answer 1

This is a system that underwent an enthalpy change. Typically we relate the data you set forth via,

`q = mC_sDeltaT`

Note: I will assume the density of water is roughly `(1g)/(cm^3)`. Unless this is extremely precise work for a more advanced course, this won't make much of a difference.

Hence,

`71*10^3J = 267g * (4.18J)/(g*K) * (T_f - 298K)`
`therefore T_f = 362K`

To be sure, in some work, we would treat the sodium and the water as separate systems equilibrating. However, you included the enthalpy change of the entire system, and you weren't given the heat capacity of sodium, so the way above may be the most effective way given that data on an exam.