1. Nov 5, 2009

### sowmit

1. The problem statement, all variables and given/known data
A 70-kg log falls from a height of 25m into a lake. If the log, the lake, and the air are all at 300k, find the change in entropy of the Universe during this process.

2. Relevant equations

delta S = Q/T

3. The attempt at a solution
T= 300k
M log=70 kg = 7000g
Specific heat of wood = .420 j/gk
h= 25 m

I found the energy for the log Q= 882000j
Entropy for it delta S = q/t = 29400j/k

Then I don't know what else to do. Someone please help. I really NEED the HELP to solve this problem.

2. Nov 5, 2009

3. Nov 5, 2009

### Andrew Mason

Ask yourself where the heat flows to when the log hits the water. Does it flow out of the log? Does heat flow into the lake? Does heat flow into the air? does heat flow into the log? Do you have to know how much flows into each to answer the problem (hint: there is a reason all three are set at the same temperature).

If there is no heat flowing out and only heat flowing in, what is the change in entropy of the log + water + air?

Note: this has nothing to do with heat capacity. I don't know where you get a Q of 882000J. Where does the energy that is turned into heat come from?

AM

4. Jan 11, 2010

### sweetdreamzzz

The potential energy lost by the log is carried away by heat, that is, the
kinetic energy of the impact into the lake get converted into heat, so
Q = (PE) = mgy = mgh = (70 kg)(9.80 m/s2)(25 m) = 1.7×10^4 J .
We will ignore air friction here and since the temperature remains constant,
we just have one term in the entropy equation. As such, the change
of entropy is
S = Q/T= 1.7 × 10^4 J / 300 K = 57 J/K .

website: http://www.etsu.edu/physics/lutter/courses/phys2010/p2010chap14.pdf