Question

A block of copper at a temperature of $\mathbf{50}\mathbf{°}\mathbf{}\mathbf{C}$ is placed in contact with a block of aluminium at a temperature of$\mathbf{45}\mathbf{°}\mathbf{}\mathbf{C}$ in an insulated container. As a result of a transfer of 2500 J of energy from the copper to the aluminium, the final equilibrium temperature of the two blocks is $\mathbf{48}\mathbf{°}\mathbf{}\mathbf{C}$. (a) What is the approximate change in the entropy of the aluminium block? (b) What is the approximate change in the entropy of the copper block? (c) What is the approximate change in the entropy of the Universe? (d) What is the change in the energy of the Universe?

Short answer

$$\begin{gathered} \mathrm{a})7.86\hspace{0.33em}\mathrm{J}/\mathrm{K} \\ \mathrm{b})-7.74\hspace{0.33em}\mathrm{J}/\mathrm{K} \\ \mathrm{c})0.12\mathrm{J}/\mathrm{K} \\ \mathrm{d})0 \end{gathered}$$

Step-by-step solution

Identification of the given data

The given data can be listed below as-

• The temperature of the copper block is${50}^{\circ }\mathrm{C}$.
• The temperature of the aluminium block is${45}^{\circ }\mathrm{C}$.
• The transfer of energy or the energy lost is 2500 J.
• The final equilibrium temperature is${48}^{\circ }\mathrm{C}$ .

Significance of the entropy

The entropy is described as the amount of the measure of the amount of the particular energy that is unavailable for doing work.

The equation of entropy gives the change in the entropy of the copper, aluminium and universe.

The change in entropy of the aluminium

a)

The equation of the change in the entropy for the aluminium is,

$$\begin{gathered} \frac{1}{\mathrm{T}}=\frac{∆{\mathrm{S}}_{\mathrm{al}}}{∆\mathrm{E}} \\ ∆{\mathrm{S}}_{\mathrm{al}}=\frac{∆\mathrm{E}}{\mathrm{T}} \end{gathered}$$

Here, $∆S_{al}$is the change in the entropy of the aluminium block,$∆E$ is the transfer of energy and T is the temperature of the aluminium block

Substituting the values in the above equation,

role="math" localid="1657877743769" $$\begin{gathered} ∆{\mathrm{S}}_{\mathrm{al}}=\frac{2500\mathrm{J}}{\left(45+273\right)\mathrm{K}} \\ =7.86\mathrm{J}/\mathrm{K} \end{gathered}$$

Thus, the change in entropy of the aluminium is $7.86\mathrm{J}/\mathrm{K}$.

The change in entropy of the copper,

b)

The equation of the change in the entropy for the copper is,

$$\begin{gathered} \frac{1}{\mathrm{T}}=\frac{∆{\mathrm{S}}_{copper}}{∆\mathrm{E}} \\ ∆{\mathrm{S}}_{copper}=\frac{∆\mathrm{E}}{\mathrm{T}} \end{gathered}$$

Here, $\Delta S_{copper}$is the change in the entropy of the copper block, $∆E$is the transfer of energy which is lost energy for copper and T is the temperature of the copper block

Substituting the values in the above equation,

role="math" localid="1657877805172" $$\begin{gathered} ∆{\mathrm{S}}_{\infty }=\frac{-2500\mathrm{J}}{\left(50+273\right)\mathrm{K}} \\ =-7.74\hspace{0.33em}\mathrm{J}/\mathrm{K} \end{gathered}$$

Thus, the change in entropy of the copper is $-7.74\hspace{0.33em}\mathrm{J}/\mathrm{K}$.

The entropy change in the universe

c)

The equation for the total entropy change of the universe is,

$S_{univ}=S_{al}+S_{co}$

Here,$S_{univ}$ is the total entropy change of the universe$S_{al}$ and $S_{co}$are the entropy of the copper and the aluminium respectively.

Substituting the values in the above equation,

$$\begin{gathered} {\mathrm{S}}_{\mathrm{univ}}=7.86\mathrm{J}/\mathrm{K}-7.74\mathrm{J}/\mathrm{K} \\ =0.12\mathrm{J}/\mathrm{K} \end{gathered}$$

Thus, the total entropy change of the universe is $0.12\mathrm{J}/\mathrm{K}$.

Step 6:The energy change in the universe

d)

As there is only the entropy change happened, thus there is no noticeable heat gained or lost by the universe. So, the change in the energy of the universe is zero.

Thus, the change in the energy of the universe is 0.