## Related questions with answers

A block of copper at a temperature of $50^{\circ} \mathrm{C}$ is placed in contact with a block of aluminum at a temperature of $45^{\circ} \mathrm{C}$ 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 $48^{\circ} \mathrm{C}.$ What is the approximate change in the entropy of the aluminum block?

Solution

VerifiedGiven: $t_{\text{Al}} = 45^{\text{\textdegree}}$C, $T_{\text{eq}} = 45 + 273 = 318$ K.

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

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

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