Creative Commons License Copyright © Michael Richmond. This work is licensed under a Creative Commons License.

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


This page maintained by Michael Richmond. Last modified Dec 19, 1997.

Creative Commons License Copyright © Michael Richmond. This work is licensed under a Creative Commons License.