Calculating Entropy Change of Lake from Thrown Aluminum Bar

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SUMMARY

The discussion focuses on calculating the entropy change (ΔS) of a lake when a 2.00 kg aluminum bar at 300 degrees Celsius is thrown into it. The specific heat capacity of aluminum is 900 J/(kg·K), and the heat absorbed by the lake is calculated to be 5.13 x 10^5 J. The formula used for calculating the entropy change is ΔS = ΔQ/T, where T is the absolute temperature of the lake, which is 15.0 degrees Celsius or 288.15 K. The final entropy change of the lake can be computed using these values.

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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.
 
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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.
 

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