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A block of copper at a temperature of 50C50^{\circ} \mathrm{C} is placed in contact with a block of aluminum at a temperature of 45C45^{\circ} \mathrm{C} in an insulated container. As a result of a transfer of 2500 J of energy from the copper to the aluminum, the final equilibrium temperature of the two blocks is 48C.48^{\circ} \mathrm{C}. What is the approximate change in the entropy of the aluminum block?

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Given: tAl=45°t_{\text{Al}} = 45^{\text{\textdegree}}C, Teq=45+273=318T_{\text{eq}} = 45 + 273 = 318 K.

Since aluminium block received energy in total of: ΔE=2500\Delta E = 2500 J, its change in entropy is:

1T=ΔSΔEΔS=ΔET=2500318ΔS=7.86J/K.\begin{align*} \frac{1}{T} &= \frac{\Delta S}{\Delta E} \\ \Delta S &= \frac{\Delta E}{T} = \frac{2500}{318} \\ \Delta S &= 7.86 \, \text{J/K.} \end{align*}

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