Entropy change of a falling object I have question which essentially asks the entropy change of a sword of initial temperature $593\,\mathrm{K}$, mass $1.2\,\mathrm{kg}$ and S.H.C of $0.13\,\mathrm{J}\cdot\mathrm{K}^{-1}\cdot\mathrm{g}^{-1}$, which falls through $1200\,\mathrm{m}$ into a very large lake of temperature $278\,\mathrm{K}$. I have been trying to solve it myself without any success, so any help would be appreciated
 A: 
The height is 1200 metres above the lake. So far I have calculated the
heat capacity of the 1.2kg sword, and tried to use the equation dS =
(CdT)/T to find the change in entropty, but with no success

Your equation is correct for a differential change (decrease) in entropy of the sword due to cooling by the lake. Just take the integral and evaluate it between the initial and final temperatures of the sword.
But to get the total entropy change (sword plus lake) due to heat transfer you also have to calculate the change in entropy of the lake due to the heat transfer to it. Here the entropy change (increase) of the lake is simply the heat transfer divided by the lake temperature since the lake is a thermal reservoir.
The entropy of the lake is also increased by the irreversible negative work (due to friction drag) it does on the sword to reduce its speed (and kinetic energy) as it goes to the bottom. Per the work energy theorem that work equals the change in kinetic energy of the sword. Here you can consider heat transfer to the lake to be the equivalent to the irreversible work done on the sword. The  trouble is the terminal velocity is not given, so one needs assume it’s negligible compared to the velocity at impact to solve problem.
Hope this helps.
A: The heat generated from the fall will be absorbed into the lake (reservoir) and therefore is not of interest in the problem.
Using the Thermodynamical Identity: $dU = T dS-pd V+\mu dN$
As pressure, volume and number of particles are fixed the relation between entropy, energy and temperature is simple:
$dU = TdS \Rightarrow dS = dU/T$
$ U = C_p * m * T \Rightarrow dU = C_p * m * dT$
$\Delta S = S(T_2)-S(T_1)= \int_{T_1}^{T_2} dS = \int_{T_1}^{T_2} 1/T  dU = \int_{593K}^{278K} C_p*m/T  dT $
$\Delta S = 0.13*1200(\ln(278/593)) J/K =  -118,2 J/K$
