For a class demonstration, your physics instructor pours \(1.00 \mathrm{~kg}\) of steam at \(100.0^{\circ} \mathrm{C}\) over \(4.00 \mathrm{~kg}\) of ice at \(0.00^{\circ} \mathrm{C}\) and allows the system to reach equilibrium. He is then going to measure the temperature of the system. While the system reaches equilibrium, you are given the latent heats of ice and steam and the specific heat of water: \(L_{\text {ice }}=3.33 \cdot 10^{5} \mathrm{~J} / \mathrm{kg}\), \(L_{\text {steam }}=2.26 \cdot 10^{6} \mathrm{~J} / \mathrm{kg}, c_{\text {water }}=4186 \mathrm{~J} /\left(\mathrm{kg}^{\circ} \mathrm{C}\right) .\) You are asked to calculate the final equilibrium temperature of the system. What value do you find?

To find the heat lost by the steam, we first need to calculate the heat required to condense the steam into water. This can be done using the latent heat of steam: \(Q_1 = m_{steam} \cdot L_{steam}\) Now, plug in the values: \(Q_1 = 1.00 \mathrm{~kg} \cdot 2.26 \cdot 10^{6} \mathrm{~J} / \mathrm{kg} = 2.26 \cdot 10^{6} \mathrm{~J}\)

To find the heat gained by the ice, we first need to calculate the heat required to melt the ice into water. This can be done using the latent heat of ice: \(Q_2 = m_{ice} \cdot L_{ice}\) Now, plug in the values: \(Q_2 = 4.00 \mathrm{~kg} \cdot 3.33 \cdot 10^{5} \mathrm{~J} / \mathrm{kg} = 1.33 \cdot 10^{6} \mathrm{~J}\)

Compare the values of \(Q_1\) and \(Q_2\): If \(Q_1 > Q_2\), then all ice melts and the final state is water only. In this case, we will then calculate the equilibrium temperature assuming all ice has melted. If \(Q_1 < Q_2\), then all steam condenses and the final state is a mixture of water and ice. In this case, the equilibrium temperature will be \(0^{\circ}C\). Here, \(Q_1 = 2.26 \cdot 10^{6} \mathrm{~J} > Q_2 = 1.33 \cdot 10^{6} \mathrm{~J}\), which means all the ice melts and we have only water in the final state.

Since all ice has melted, we will now calculate the heat required to raise the temperature of the water formed from ice: \(Q_3 = m_{ice} \cdot c_{water} \cdot \Delta T_{ice}\) And calculate the heat required to lower the temperature of the water formed from steam: \(Q_4 = m_{steam} \cdot c_{water} \cdot \Delta T_{steam}\) We set both heat transfer values to be equal, as the heat lost by the steam-water must be gained by the ice-water: \(Q_3 = Q_4 \Rightarrow m_{ice} \cdot c_{water} \cdot \Delta T_{ice} = m_{steam} \cdot c_{water} \cdot \Delta T_{steam}\) The equilibrium temperature can then be calculated by solving for the temperature difference: \(\Delta T_{eq} = \frac{m_{steam} \cdot \Delta T_{steam}}{m_{ice}}\) We know the initial temperatures of steam and ice, so we can find the temperature differences: \(\Delta T_{steam} = 100^{\circ} \mathrm{C} - T_{eq}\) \(\Delta T_{ice} = T_{eq} - 0^{\circ} \mathrm{C}\) Substituting the initial temperature values and masses of steam and ice, we find the equilibrium temperature: \(T_{eq} = \frac{1.00 \mathrm{~kg}\cdot(100^{\circ} \mathrm{C} - T_{eq})}{4.00 \mathrm{~kg}}\) Solve for \(T_{eq}\): \(T_{eq} = \frac{100}{5} = 20 ^{\circ} \mathrm{C}\) The final equilibrium temperature of the system is \(20^{\circ} \mathrm{C}\).