You're on a roll — Here's a calculation problem involving temperature and thermal equilibrium: A $200 \text{ g}$ copper block at $100^\circ\text{C}$ is placed into $300 \text{ g}$ of water at $20^\circ\text{C}$ in an insulated container. Calculate the final equilibrium temperature of the system. Given: Specific heat capacity of copper ($c_{Cu}$) = $385 \text{ J/(kg}\cdot^\circ\text{C)}$ Specific heat capacity of water ($c_w$) = $4186 \text{ J/(kg}\cdot^\circ\text{C)}$ Step 1: Convert masses to kilograms. $$m_{Cu} = 200 \text{ g} = 0.200 \text{ kg}$$ $$m_w = 300 \text{ g} = 0.300 \text{ kg}$$ Step 2: Apply the principle of thermal equilibrium, which states that the heat lost by the hotter object (copper) equals the heat gained by the cooler object (water). $$Q_{lost, Cu} = Q_{gained, w}$$ The formula for heat transfer is $Q = mc\Delta T$. Let $T_f$ be the final equilibrium temperature. $$m_{Cu} c_{Cu} (T_{Cu,i} - T_f) = m_w c_w (T_f - T_{w,i})$$ Step 3: Substitute the given values into the equation. $$(0.200 \text{ kg})(385 \text{ J/(kg}\cdot^\circ\text{C)})(100^\circ\text{C} - T_f) = (0.300 \text{ kg})(4186 \text{ J/(kg}\cdot^\circ\text{C)})(T_f - 20^\circ\text{C})$$ Step 4: Simplify and solve for $T_f$. $$77 (100 - T_f) = 1255.8 (T_f - 20)$$ $$7700 - 77 T_f = 1255.8 T_f - 25116$$ $$7700 + 25116 = 1255.8 T_f + 77 T_f$$ $$32816 = 1332.8 T_f$$ $$T_f = \frac{32816}{1332.8}$$ $$T_f \approx 24.62 \text{ }^\circ\text{C}$$ The final equilibrium temperature of the system is approximately $24.62^\circ\text{C}$. $$\boxed{24.62^\circ\text{C}}$$