Physics 150  Fall 2003

Solution to HW 4


1.   You should write a short essay including the points:

Conservation laws define a relation that a sum of quantities (each measurable at least in principle) is conserved, i.e., never changes in an isolated system.  They are useful because they allow important, useful consequences to be derived without knowing all the details.

For the second part of the question, there can be different answers and the grade depends on how well you justify your answers.  My belief is that the laws are an invention humans (like other scientific laws); nevertheless, they say something definite about nature to the extent that they are tested by experiment and are found to describe nature.   This is exactly like other laws, but the conservation laws are more global and make possible greater generalizations that must be tested by experiments.


2.  The dye disperses through the water.  This is a visible example that is just like heat dispersing, etc.  Since the molecules are free to move in the water, it is more probable to find the molecules mixed with water molecules than all clumped together, which is very unlikely. 


3.  Your answer is graded according to the way the arguments are supported and the correctness within the laws of science.  The first part of the question is “non-scientific”: I do not know the exact reason for Morris’ choice of arguments, but many people use arguments based upon scientific laws because it appears to make their case stronger  whether or not the law is used correctly.  Certainly evolution into organisms with greater organization represents an increase in order.  This does not violate the second law if the increase in order occurs in a system that is not isolated; the input of energy at a higher temperature than the output is consistent with useful work being done and a increase of order in that system.  The universe as a whole has a decrease of order


4.  Using energy conservation, ½ mv2 = mgh or  h =  ½ mv2/g.  In this case,

h= ½ (10m/s) 2/10 m2/s = 5m.


5.  Using momentum conservation, m1v1 + m2v2 = 0 or v2 = -m1v1/ m2. In this case,

v2 = -(90/40) 1.5 m/s = 3.375 m/s.


6.  Using energy conservation, ½ mv2 + mgh + H = constant, where H = heat energy. In this case v=0 at the start and finish, so that H = mg(hstart – h finish) = 800 Kg * 10 m2/s * 18m = 144,000 J.


7.  Using energy conservation, ½ mv2 + H = constant, so H = -½ m (v2 final - v2 initial ) =
½ m v2 initial = 0.5 * 0.25 Kg * (50 m/s) 2 = 312.5 J = 74.7 calories.  The change in temperature is  Delta T = H/c m.  In this case c = 1 cal/K gram and m = 300 gram, so

Delta T =  74.7 cal/ (1 cal/K gram * 300 gram) = 0.25 K = 0.25 degrees centigrade.


8.  A power of 1000W for 1 minute is an energy of 1000 J/s * 60 s = 60,000 J = 14340 cal. Since 5 L of water has mass 5 Kg, it follows that to increase the temperature by 80 degrees, one must supply an energy of H = 1 cal/K gram * 80 K * 5000 gram = 400,000 cal.  This requires 400,000/14,340 = 27.9 minutes.


9.  The solution to part of the problem is given by Lightman: work = mgh = 5 Kg * 10 m2/s * 0.3 m = 15 J = 3.59 cal.  The heat input is 300K * 0.02 Kg * 655 J/Kg K = 3930 J = 939 cal.  Almost all the energy goes into heating the gas; very little goes into work in this example.   (This is a case with a great “mismatch”.  The efficiency would be much greater if the same mass of gas was used to raise a larger mass.)


10.  The maximum possible efficiency of any heat enegine is e = 1 – Tout/Tin , where the temperatures are in Kelvin.  There is a lower limit to Tout in practice.  For example, steam engines need to exhaust the steam at T greater than 100 C.  Thus the way to increase the efficiency is to increase Tin .  This is what is actually done, which requires development of engines  that can withstand the high input temperatures.

In the example given, Tin = 100 + 273 = 373K and  Tout = 1273K, so that the maximum efficiency is 1 – 373 / 1273  = 71%.