Question

Kerepakupai Vena in Venezuela, more commonly known as Angel Falls, is the world's highest waterfall with a height of 979 m and 47 different plunges. The longest plunge (for the water) is 807 m. At the bottom of the fall, the kinetic energy of the water is converted into thermal energy. This results in a temperature increase of the water. If the water has a temperature of 17.1°C when it gets to the bottom of the plunge, what was the temperature at the top of the plunge? The specific heat of water is 1.00 kcal/(kg · °C).

Answer 1

Answer:
`15.21 \°C`

Step-by-step explanation:

We will use the the conservation of energy principle to solve this problem. In this sense, for the case of the Kerepakupai Vena waterfall the energy at the top must be equal to the energy at the bottom:

`E_(top)=E_(bottom)` (1)

`E_(top)=K_(top)+U_(top)` (2)

`E_(bottom)=K_(bottom)+U_(bottom)` (3)

Where
`K_(top)` and
`U_(top)` are the kinetic an potential energy at the top of the waterfall, respectively; and
`K_(bottom)` and
`U_(bottom)` are the kinetic an potential energy at the bottom of the waterfall, respectively.

Since we are told the kinetic energy of the water is converted into thermal energy
`Q`, we have:

`K_(top)=Q_(top)` and
`K_(bottom)=Q_(bottom)`

Hence (2) and (3) are rewritten as:

`E_(top)=Q_(top)+U_(top)` (4)

`E_(bottom)=Q_(bottom)+U_(bottom)` (5)

Then:

`Q_(top)+U_(top)=Q_(bottom)+U_(bottom)` (6)

On the other hand we know the potential energy for both cases is:

`U_(top)=mgh_(top)` and
`U_(bottom)=mgh_(bottom)`

Where:

`m` is the mass of water

`g=9.8 m/s^(2)` is the acceleration due gravity

`h_(top)=807 m` is the height at the top

`h_(bottom)=0 m` is the height at the bottom of the waterfall

In addition, the thermal energy for both cases is:

`Q_(top)=m. c. T_(top)` and
`Q_(bottom)=m. c. T_(bottom)`

Where:

`c=1 (kcal)/(kg \°C)=4.186(10)^(3) (J)/(kg \°C)` is the specific of water

`T_(top)=` is the temperature at the top

`T_(bottom)=17.1 \°C` is the temperature at the bottom

So, keeping this in mind, equation (6) is rewritten as:

`m. c. T_(top)+mgh_(top)=m. c. T_(bottom)+mgh_(bottom)` (7)

Since
`h_(bottom)=0 m`:

`m. c. T_(top)+mgh_(top)=m. c. T_(bottom)` (8)

Finding
`T_(bottom)-T_(top)= \Delta T`:

`T_(bottom)-T_(top)=\Delta T=(gh)/(c)` (9)

Solving:

`\Delta T=((9.8 m/s^(2))(807 m))/(4.186(10)^(3) (J)/(kg \°C))` (10)

`\Delta T=1.889 \°C` (11)

Now that we have
`\Delta T` and
`T_(bottom)` we can finally find
`T_(top)`:

`T_(top)=\Delta T-T_(bottom)` (12)

`T_(top)=1.889 \°C-17.1 \°C` (13)

Therefore:

`T_(top)=15.21 \°C`