A \(110 .-\mathrm{g}\) sample of copper (specific heat capacity \(=0.20 \mathrm{~J} /{ }^{\circ} \mathrm{C}\). \(\mathrm{g}\) ) is heated to \(82.4^{\circ} \mathrm{C}\) and then placed in a container of water at \(22.3^{\circ} \mathrm{C}\). The final temperature of the water and copper is \(24.9^{\circ} \mathrm{C}\). What is the mass of the water in the container, assuming that all the heat lost by the copper is gained by the water?

From the exercise, we know the following: - The mass of copper (m_copper) = 110 g - The specific heat capacity of copper (c_copper) = 0.20 J/g°C - The initial temperature of copper (T_initial_copper) = 82.4°C - The initial temperature of water (T_initial_water) = 22.3°C - The final temperature of both copper and water (T_final_both) = 24.9°C

We know that when we put the hot copper inside the container, the copper will lose heat and the water will gain heat until both reach the same final temperature. So, we can use the following equation for heat transfer when all the heat lost by the copper is gained by the water: \(Q_{lost\_copper} = Q_{gained\_water}\)

We can find the heat lost by the copper using the equation: \(Q_{lost\_copper} = m_{copper} \times c_{copper} \times \Delta T_{copper}\) Where: - m_copper = mass of copper - c_copper = specific heat capacity of copper - ∆T_copper = change in temperature of copper = |T_final_both - T_initial_copper| Now, substituting the values, we get: \(Q_{lost\_copper} = (110) \times (0.20) \times (|24.9 - 82.4|)\) Calculating the heat lost by copper: \(Q_{lost\_copper} = 110 \times 0.20 \times 57.5 = 1265\ \mathrm{J}\)

Now, we will find the mass of water using the heat gained by water: \(Q_{gained\_water} = m_{water} \times c_{water} \times \Delta T_{water}\) Where: - m_water = mass of water - c_water = specific heat capacity of water = 4.18 J/g°C (given) - ∆T_water = change in temperature of water = |T_final_both - T_initial_water| Since we know that \(Q_{lost\_copper} = Q_{gained\_water}\), we can rewrite the equation as: \(m_{water} = \frac{Q_{gained\_water}}{c_{water} \times \Delta T_{water}}\) Substituting the values, we get: \(m_{water} = \frac{1265}{(4.18) \times (|24.9 - 22.3|)}\) Calculating the mass of water: \(m_{water} = \frac{1265}{4.18 \times 2.6} = 72.5\ \mathrm{g}\) So, the mass of the water in the container is 72.5 g.