To solve this problem it is necessary to apply the concepts concerning the conservation of both potential and thermodynamic energy of the body. That is to say that as the body has a loss of potential energy it is gained in the form of thermal energy on water. If the potential energy is defined as
[tex]PE = mgh[/tex]
Where,
m= mass
g = Gravitational acceleration
h = Height
And thermal energy is obtained as
[tex]Q = mC_p\Delta T[/tex]
Where,
[tex]\Delta T[/tex] = Change in Temperature
[tex]C_p =[/tex] Specific Heat
m = Mass
We can equate this equation and rearrange to find the change at the Temperature, then
[tex]mgh = mC_p\Delta T[/tex]
[tex]\Delta T = \frac{gh}{C_p}[/tex]
Our values are given as,
[tex]C_p = 4186J/Kg\cdot K \rightarrow[/tex] Specific Heat Water
Using energy conservation
[tex]g = 9.8m/s^2[/tex]
[tex]h = 807m[/tex]
Replacing,
[tex]\Delta T = \frac{(9.8)(807)}{4186}[/tex]
[tex]\Delta T = 1.89K[/tex]
Therefore the temperature increase in a 1kg sample of water is 1.89K
