A \(2.0 \cdot 10^{2}\) g piece of copper at a temperature of \(450 \mathrm{~K}\) and a \(1.0 \cdot 10^{2} \mathrm{~g}\) piece of aluminum at a temperature of \(2.0 \cdot 10^{2} \mathrm{~K}\) are dropped into an insulated bucket containing \(5.0 \cdot 10^{2} \mathrm{~g}\) of water at \(280 \mathrm{~K}\). What is the equilibrium temperature of the mixture?

The heat lost by a substance equals the heat gained by another substance in an isolated system (no heat lost to the surroundings). Mathematically, this can be expressed as: $$Q_{lost} = Q_{gained}$$ or $$mc \Delta T = mc \Delta T$$ where, m: mass of the substance c: specific heat capacity of the substance ∆T: change in temperature of the substance

We will need specific heat capacities for copper (c_c), aluminum (c_a), and water (c_w) to solve for the equilibrium temperature. These values are: c_c = 386 J/kg·K (for copper) c_a = 897 J/kg·K (for aluminum) c_w = 4186 J/kg·K (for water)

Converting mass from grams to kilograms (m_c for copper, m_a for aluminum, and m_w for water) makes it easier to calculate heat exchanges. m_c = 2.0 * 10^2 g = 0.2 kg m_a = 1.0 * 10^2 g = 0.1 kg m_w = 5.0 * 10^2 g = 0.5 kg

Following conservation of energy, we can set up the following equation: $$m_c c_c (T_{final} - T_{c}) + m_a c_a (T_{final} - T_{a}) = m_w c_w (T_{w} - T_{final})$$ where, T_c = initial temperature of copper T_a = initial temperature of aluminum T_w = initial temperature of water T_final = equilibrium temperature

The given initial temperatures for copper, aluminum, and water are: T_c = 450 K T_a = 2.0 * 10^2 K = 200 K T_w = 280 K Substituting these values into the heat exchange equation we derived earlier: $$0.2 \cdot 386 (T_{final} - 450) + 0.1 \cdot 897 (T_{final} - 200) = 0.5 \cdot 4186 (280 - T_{final})$$ Solve this equation to find the equilibrium temperature (T_final): $$-77.2 T_{final} + 34740 + 89.7 T_{final} - 17940 = -2093 T_{final} + 585620$$ Combine the like terms and simplify the equation: $$2005.5 T_{final} = 544420$$ Finally, divide by 2005.5 to find the equilibrium temperature: $$T_{final} = \frac{544420}{2005.5} \approx 271.3 K$$ The equilibrium temperature of the mixture is approximately 271.3 K.