Answer:
[tex]\Delta T=0.14^{\circ}C[/tex]
Explanation:
If we consider that all the gravitational potential energy ([tex]U=mgh[/tex]) gets transformed into thermal energy ([tex]Q=mc\Delta T[/tex]), we then have:
[tex]mgh=mc\Delta T[/tex]
which means:
[tex]\Delta T=\frac{gh}{C}[/tex]
Since the specific heat capacity of water is [tex]c=4186 J/kg^{\circ}K[/tex], we have:
[tex]\Delta T=\frac{(9.8m/s2)(60m)}{4186J/kg^{\circ}K}=0.14^{\circ}K=0.14^{\circ}C[/tex]
(A difference in Kelvin is the same difference is Celsius).
