A large block of copper initially at \(20.0^{\circ} \mathrm{C}\) is placed in a vat of hot water \(\left(80.0^{\circ} \mathrm{C}\right) .\) For the first $1.0 \mathrm{J}$ of heat that flows from the water into the block, find (a) the entropy change of the block, (b) the entropy change of the water, and (c) the entropy change of the universe. Note that the temperatures of the block and water are essentially unchanged by the flow of only \(1.0 \mathrm{J}\) of heat.

First, we will calculate the entropy change of the copper block when it gains \(1.0\,\text{J}\) of heat. The temperature of the block will remain constant at \(20.0^{\circ} \mathrm{C}\). We need to convert the given temperature of the block to Kelvin by adding 273.15 to the Celsius temperature. $$ T_\mathrm{block} =20.0^{\circ} \mathrm{C} + 273.15 = 293.15\,\mathrm{K} $$ Now, we can calculate the entropy change using the equation for entropy change: $$ \Delta S_\mathrm{block}=\frac{Q}{T_\mathrm{block}} $$ Substitute the values: $$ \Delta S_\mathrm{block}=\frac{1.0\,\text{J}}{293.15\,\mathrm{K}} $$ $$ \Delta S_\mathrm{block}=3.41\times10^{-3}\,\text{J/K} $$ So, the entropy change of the block is \(3.41\times10^{-3}\,\text{J/K}\).

Now, we will calculate the entropy change of the water when it loses \(1.0\,\text{J}\) of heat. The temperature of the water will remain constant at \(80.0^{\circ} \mathrm{C}\). We need to convert the given temperature of the water to Kelvin by adding 273.15 to the Celsius temperature. $$ T_\mathrm{water} =80.0^{\circ} \mathrm{C} + 273.15 = 353.15\,\mathrm{K} $$ Now, the water loses \(1.0\,\text{J}\) of heat, so we put a negative sign to the \(Q\). $$ \Delta S_\mathrm{water}=\frac{-Q}{T_\mathrm{water}} $$ Substitute the values: $$ \Delta S_\mathrm{water}=\frac{-1.0\,\text{J}}{353.15\,\mathrm{K}} $$ $$ \Delta S_\mathrm{water}=-2.83\times10^{-3}\,\text{J/K} $$ So, the entropy change of the water is \(-2.83\times10^{-3}\,\text{J/K}\).

Now, we will calculate the total entropy change of the universe, which is the sum of the entropy change of the block and the entropy change of the water. $$ \Delta S_\mathrm{universe}=\Delta S_\mathrm{block}+\Delta S_\mathrm{water} $$ Substitute the values: $$ \Delta S_\mathrm{universe}=3.41\times10^{-3}\,\text{J/K} + (-2.83\times10^{-3}\,\text{J/K}) $$ $$ \Delta S_\mathrm{universe}=5.80\times10^{-4}\,\text{J/K} $$ So, the entropy change of the universe is \(5.80\times10^{-4}\,\text{J/K}\). In conclusion, the entropy change of the block is \(3.41\times10^{-3}\,\text{J/K}\), the entropy change of the water is \(-2.83\times10^{-3}\,\text{J/K}\), and the entropy change of the universe is \(5.80\times10^{-4}\,\text{J/K}\).