Question

A waterfall is 60 meters high. What is the temperature rise in Kelvins of the water from just before to just after it hits the rocks at the bottom of the falls, assuming negligible air resistance during the fall and that the water doesn't rebound but just splats onto the rock

Answer 1

`\Delta T=0.14^(\circ)C`

Step-by-step explanation:

If we consider that all the gravitational potential energy (
`U=mgh`) gets transformed into thermal energy (
`Q=mc\Delta T`), we then have:

`mgh=mc\Delta T`

which means:

`\Delta T=(gh)/(C)`

Since the specific heat capacity of water is
`c=4186 J/kg^(\circ)K`, we have:

`\Delta T=((9.8m/s2)(60m))/(4186J/kg^(\circ)K)=0.14^(\circ)K=0.14^(\circ)C`

(A difference in Kelvin is the same difference is Celsius).