Calculating Entropy Change of Lake from Thrown Aluminum Bar

[swimchick]

An aluminum bar of mass 2.00 kg at 300 degrees C is thrown into a lake. The temperature of the water in the lake is 15.0 degrees C; the specific heat capacity of aluminum is 900 J/(kg* K).

The bar eventually reaches thermal equilibrium with the lake. What is the entropy change DeltaS_lake of the lake? Assume that the lake is so large that its temperature remains virtually constant.
Express your answer numerically in joules per kelvin.

The change in Q of the Lake, heat absorbed by the lake is = 5.13*10^5 J

When there is a heat transfer of change in Q to a substance at constant temperature T, the entropy change DeltaS of the substance is given by
\Delta S={\Delta Q}/{T},
where T is absolute temperature.

 

[Chi Meson]

Where's the problem? You have the Q, you can get the T (temp in Kelvins of the lake) and you have the equation ∆S=Q/T.

 

[Moetasim]

In this case, the lake is assumed to be so large that its temperature remains virtually constant, so the temperature T can be considered constant. Therefore, the entropy change of the lake can be calculated as:

DeltaS_lake = (5.13*10^5 J) / (15.0 degrees C + 273.15 K) = 1681.35 J/K

This means that the entropy of the lake has increased by 1681.35 J/K due to the heat transfer from the aluminum bar. This increase in entropy is a result of the randomization of the molecules in the lake as they absorb heat from the bar, leading to an increase in disorder.