Calculating Entropy Change of Lake from Thrown Aluminum Bar

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To calculate the entropy change of the lake when a 2.00 kg aluminum bar at 300 degrees C is thrown into it, the heat absorbed by the lake is determined to be 5.13 x 10^5 J. The lake's temperature is assumed constant at 15.0 degrees C, which converts to 288.15 K in absolute terms. Using the formula for entropy change, ΔS = ΔQ/T, the entropy change for the lake can be calculated. The discussion emphasizes the straightforward application of the formula once the values for heat transfer and temperature are established. The final entropy change can be expressed numerically in joules per kelvin.
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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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