1. **State the problem:** We have 500 mg (0.5 g) of iron at 200°C placed into 250 g of water at 20°C. We want to find the final temperature $T_f$ of the mixture after thermal equilibrium. 2. **Formula and principles:** The heat lost by iron equals the heat gained by water (assuming no heat loss to surroundings): $$m_{Fe} c_{Fe} (T_{Fe,i} - T_f) = m_{H_2O} c_{H_2O} (T_f - T_{H_2O,i})$$ where: - $m_{Fe} = 0.5$ g - $c_{Fe} = 0.449$ J/g°C (specific heat capacity of iron) - $T_{Fe,i} = 200$°C - $m_{H_2O} = 250$ g - $c_{H_2O} = 4.18$ J/g°C (specific heat capacity of water) - $T_{H_2O,i} = 20$°C 3. **Set up the equation:** $$0.5 \times 0.449 \times (200 - T_f) = 250 \times 4.18 \times (T_f - 20)$$ 4. **Simplify both sides:** $$0.2245 (200 - T_f) = 1045 (T_f - 20)$$ 5. **Distribute:** $$44.9 - 0.2245 T_f = 1045 T_f - 20900$$ 6. **Bring all terms involving $T_f$ to one side and constants to the other:** $$44.9 + 20900 = 1045 T_f + 0.2245 T_f$$ 7. **Combine like terms:** $$20944.9 = 1045.2245 T_f$$ 8. **Solve for $T_f$:** $$T_f = \frac{20944.9}{1045.2245}$$ 9. **Calculate:** $$T_f \approx 20.03\,^{\circ}C$$ 10. **Interpretation:** The final temperature is approximately 20.03°C, very close to the initial water temperature because the iron mass is very small compared to water.