A 5 kg ball of clay falls from a height of 6.5 meters and strikes the ground. 92% of the ball's total energy of 319 J is converted into work distorting the clay upon impact, leaving 26 J of energy converted to heat. Using the specific heat of soil of 1050 J/kg°C, the temperature increase of the clay is calculated to be 0.005°C. While a small change, this example demonstrates that heat is genuinely produced during the conversion of potential to kinetic and work/heat energy.

1. 2 ωραίες ασκήσεις
A thermometer in a hot liquid reads 77° F. What temperature is this on the
Celsius scale and on the Kelvin scale?
TC = 5/9 (TF - 32) = 5/9 (77° - 32) = 5/9 (45°) = 25°C
TK = 25° + 273 = 25° + 273 = 298 K
Notice that we first convert to Celsius, and then to Kelvin. Also, there is no misprint in
units. The degree marking is not used in the Kelvin designation of temperature, even
though it is used in both the Fahrenheit and Celsius notations.
Heat is a form of energy, and this knowledge can help us to calculate heat loses and
gains. If an object is dropped near Earth's surface, some of its energy is converted
into heat when it strikes the ground, as shown in the following problem.
A 5.00-kg ball of clay falls from a height of 6.50 meters. If 92.0% of its energy is
converted into distortional work that flattens the clay upon impact, how much
heat is produced when the clay strikes the ground?
The total energy at the top of the fall is converted into kinetic energy just before
impact with the ground (h = 0), and the kinetic energy is transformed into work and
heat when the ball actually strikes the ground. We could, therefore, calculate the
kinetic energy from the velocity of the clay just before impact. (This velocity is 11 m/s
if you decide to do this calculation for practice.)
Because the total energy remains the same, however, it is easier to use conservation
of energy and convert directly from potential energy into heat and work. The change
in potential energy of the clay must be equal to the work done to distort the clay plus
any heat energy that the clay retains after impact.
The change in potential energy of the clay is given by:
Ep = m g h = 5.00 kg (9.80 m/s2
)(6.50 m) = 319 J
This means a total of 319 J of energy is available from the fall. The work done to
distort the clay on impact is given as 92.0%, but we must use the decimal equivalent
0.920 in our calculation. The work causing distortion is then negative because it is
work done on the clay.
W = -0.920 E = - 0.920 (319 J) = -293 J
The heat retained by the clay ball can then be found using the second law of
thermodynamics:
H = E + W = 319 J + (-293 J) = 26 J
Thus we see that most of the energy goes to distort the molecular structure of the
clay and only a small fraction of the energy causes an increase in the total internal
KE of the molecules. The joule is a perfectly acceptable unit for heat, but we can also
express the heat in kilocalories.
26 J (1 kcal / 4186 J) = 0.0062 kcal

2. How much increase in temperature does the clay in question 2 experience upon impact?
Here we mustuse the conceptof specific heatin our calculation.Since the specific heatis given in
Table 5.1 (see textbook) in both kcal/kg-C° and J/kg-C°, we can use either 26 J or 0.0062 kcal in our
determination,as long as we choose the proper specific heatvalue. Since "clay" is not given specifically,
we will use the value for "soil,"which should be close enough for our example problem.There are tables
similar to 5.1 in other books that contain specific heatvalues for almostanymaterial that you mightneed
to work with, so we could find an exact value for the specific heatof clay if we really needed to. Such
tables mightbe found in your local library.
H = m c T, but we must first solve for T
T = H / m c = 26 J / [5.00 kg (1050 J/kg-C°)] = 0.005° C
Remember that this is the change in temperature, not the final temperature of the
clay. If the clay has begun its fall with a temperature of 21.000°,the temperature ofthe clay would now
be only 21.005°,which is much too small a change to notice. We had to use 5 significantfigures in our
energy calculation to find such a small change,butheatreally is generated in this process and itis not
always negligible in real-life situations.As an example,consider the work done by the brakes to slow
down your car if you are going 55 mph and mustmake a sudden stop.Almostall of the work done to
reduce the kinetic energy of the moving car is converted by friction into heat, and the temperature of
your braking system will increase dramatically.