This was asked as an addendum to another thread so I have started it as a new thread.

Originally Posted by **casperneo** I have another question regarding thermodynamics,
A ship's anchor, made of 500kg of steel and initially at a temperature of 20 degree is dropped into the ocean which has a temperature of 7 degree. Determine the __entropy change of the anchor__.
Can i assume that since the ocean has a huge volume, the ocean water would not change temperature. Thus, the steel anchor will have a final temperature of 7 degree.
Solving it, using Cp= 4.002KJ/KG.K
The answer is -0.18162 KJ/K. |

I'm sorry but this is not correct since the temperature of the anchor varies during the cooling.

In any case you should have started a new thread for this.

I will do this for you and we can continue the discussion there.

For incompressible substances dv = 0 so combining the first law and the definition of the second law entropy

In differential form

ds = du/T = Cdt/T

So for a temperature change from T1 to T2

The entropy change S2 - S1 =[ Integral CdT/T ] from T1 to T2

If we can take the specific heat as constant then

S2 - S1 = C ln T2/T1, after performing the integration.

I think this is a bit above high school level though.