Here's another calculation problem on temperature and thermal equilibrium: A $150 \text{ g}$ aluminum block at $150^\circ\text{C}$ is submerged into $400 \text{ g}$ of water at $25^\circ\text{C}$ in an insulated calorimeter. Determine the final equilibrium temperature of the mixture. Given: Specific heat capacity of aluminum ($c_{Al}$) = $900 \text{ J/(kg}\cdot^\circ\text{C)}$ Specific heat capacity of water ($c_w$) = $4186 \text{ J/(kg}\cdot^\circ\text{C)}$ Step 1: Convert masses to kilograms. $$m_{Al} = 150 \text{ g} = 0.150 \text{ kg}$$ $$m_w = 400 \text{ g} = 0.400 \text{ kg}$$ Step 2: Apply the principle of thermal equilibrium, where the heat lost by the aluminum equals the heat gained by the water. $$Q_{lost, Al} = Q_{gained, w}$$ Using the formula $Q = mc\Delta T$, let $T_f$ be the final equilibrium temperature. $$m_{Al} c_{Al} (T_{Al,i} - T_f) = m_w c_w (T_f - T_{w,i})$$ Step 3: Substitute the given values into the equation. $$(0.150 \text{ kg})(900 \text{ J/(kg}\cdot^\circ\text{C)})(150^\circ\text{C} - T_f) = (0.400 \text{ kg})(4186 \text{ J/(kg}\cdot^\circ\text{C)})(T_f - 25^\circ\text{C})$$ Step 4: Simplify and solve for $T_f$. $$135 (150 - T_f) = 1674.4 (T_f - 25)$$ $$20250 - 135 T_f = 1674.4 T_f - 41860$$ $$20250 + 41860 = 1674.4 T_f + 135 T_f$$ $$62110 = 1809.4 T_f$$ $$T_f = \frac{62110}{1809.4}$$ $$T_f \approx 34.33 \text{ }^\circ\text{C}$$ The final equilibrium temperature of the mixture is approximately $34.33^\circ\text{C}$. $$\boxed{34.33^\circ\text{C}}$$ Last free one today — make it count tomorrow, or type /upgrade for unlimited.