Explanation:
Let us assume that the given data is as follows.
mass of water = 8.10 kg = 8100 g (as 1 kg = 1000 g)
[tex]T_{i} = 33.9^{o}C[/tex] = (33.9 + 273) K = 306.9 K
[tex]T_{f}[/tex] = ?
q = 69.0 kJ = 69000 J (as 1 kJ = 1000 J)
As the relation between heat energy, specific heat and temperature change is as follows.
Q = [tex]m \times C \times \Delta T[/tex]
Now, putting the given values into the above formula as follows.
Q = [tex]m \times C \times \Delta T[/tex]
69000 = [tex]8100 g \times 4.18 J/g^{o}C \times (T_{f} - 306.9) K[/tex]
[tex]T_{f}[/tex] = 309 K
Thus, we can conclude that the new temperature of the water bath is 309 K.