# Physics 212, Quiz #3a: Dec 19, 1997

Show all work -- you may receive partial credit!

Water in a river is at a temperature of T = 10 degrees Celsius. It then flows over a waterfall and down a height h = 50 m to the pool below.

Question 1: What is the gravitational potential energy of a cube of water 2 meters on a side, at the top of the falls? What is the kinetic energy of the same cube just before it hits the pool?

```  Answer: The volume of the cube is V = (2 m)^3 = 8 m^3.  The density
of water is rho = 1000 kg/m^3.  Therefore, the mass of the
cube is
m = (density)*(volume)
= (1000 kg/m^3)*(8 m^3)
= 8,000 kg

The gravitational potential energy of this cube is

PE = mgh = (8000 kg)*(9.8 m/s^2)*(50 m)
= 3.92 x 10^6 J

As it falls, the PE is converted to KE.  Just before it hits
the pool, all the PE has been turned into KE:

KE = 3.92 x 10^6 J
```

(If you don't know how to do problem 1, assume that the amount of kinetic energy is 1,000,000 Joules).

Assume that 25% of the water's kinetic energy is turned into heat, raising the temperature of the same 2-m cube of water.

Question 2: What is the temperature of the water at the bottom of the falls?

```  Answer: If 25% of the kinetic energy is turned into heat energy,
then the total amount of heat energy in the cube is

Heat energy = 0.25*(3.92 x 10^6 J) = 980,000 J

The heat energy goes into raising the temperature of this
cube of water:

heat energy = (specific heat)*(mass)*(change in temp)

We can solve for the change in temperature:

heat energy
change in temp  = ------------------------
(specific heat)*(mass)

980,000 J
= ------------------------
(4186 J/kg*C)*(8000 kg)

= 0.029 Celsius degrees

The temperature at the bottom of the falls is

final temp  = (initial temp) + (change in temp)
= 10 + 0.029
= 10.029 degrees C
```

Question 3: Convert the final temperature of the water in the pool from Celsius to Fahrenheit, and to Kelvin.

```  Answer:        temp K       = temp C + 273.15
= 10.029 + 273.15 = 283.30 K

temp F       = (temp C)*(9/5) + 32
= (10.029)*(1.8) + 32
= 50.05 degrees F
```