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Example: A 150 g piece of metal at 80°C is placed into 200 g of water at 20°C in a calorimeter. Find the final temperature. (Specific heat of metal = 0.385 J/g·°C, specific heat of water = 4.18 J/g·°C, assume no heat loss.)\n\nStep 1: Let the final temperature be Tf.\nStep 2: Heat lost by metal = m_metal·c_metal·(T_initial_metal – Tf).\nStep 3: Heat gained by water = m_water·c_water·(Tf – T_initial_water).\nStep 4: Set heat lost = heat gained: 150·0.385·(80 – Tf) = 200·4.18·(Tf – 20).\nStep 5: Solve: 57.75·(80 – Tf) = 836·(Tf – 20).\nStep 6: Expand: 4620 – 57.75Tf = 836Tf – 16720.\nStep 7: Bring terms together: 4620 + 16720 = 836Tf + 57.75Tf → 21340 = 893.75Tf.\nStep 8: Tf = 21340 / 893.75 ≈ 23.9°C.\n\nThus, the final equilibrium temperature is about 24°C.