Explanation:
As the given data is as follows.
m = 6.51 kg, g = 9.8 [tex]m/s^{2}[/tex], h = 66.8 m
mass of water (M) = 0.68 kg, Specific heat of water = 4200 [tex]J/kg^{o}C[/tex]
[tex]T_{1} = 15^{o}C[/tex]
According to the given situation, the decrease in potential energy of mass will be equal to heat energy gained by water.
Therefore,
mgh = [tex]m \times C \times (T_{f} - T_{i})[/tex]
[tex]6.51 kg \times 9.8 \times 66.8 m = 0.68 kg \times 4200 J/kg^{o}C \times (T_{f} - 15)^{o}C[/tex]
4261.7064 = [tex]2856 (T_{f} - 15)^{o}C[/tex]
1.492 = [tex]T_{f}[/tex] - 15
[tex]T_{f} = 16.492^{o}C[/tex]
So, [tex]\Delta T = (16.492 - 15)^{o}C[/tex]
= [tex]1.492^{o}C[/tex]
Therefore, we can conclude that rise in temperature will be [tex]1.492^{o}C[/tex].
