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Problem: 100 grams of ice at 0°C is mixed with 100 grams of water at 100°C. Find the final temperature of the mixture, assuming no heat is lost to the surroundings. (Specific heat of water = 4.2 J/g°C, Latent heat of fusion of ice = 334 J/g). Step 1: Calculate heat required to melt the ice. Heat needed = mass x latent heat = 100g x 334 J/g = 33,400 Joules. Step 2: Calculate heat released by the hot water cooling down to 0°C. Heat released = mass x specific heat x change in temp = 100g x 4.2 J/g°C x (100-0)°C = 42,000 Joules. Step 3: Compare the heats. The hot water can provide 42,000 J, but the ice only needs 33,400 J to melt. Since 42,000 > 33,400, all the ice will melt. Step 4: Calculate the remaining heat. Remaining heat = 42,000 - 33,400 = 8,600 Joules. Step 5: This remaining heat warms up the total water (100g from melted ice + 100g original water = 200g). Let the final temperature be T. Heat used = 200g x 4.2 J/g°C x T. So, 200 x 4.2 x T = 8,600. 840 x T = 8,600. T = 8,600 / 840 ≈ 10.24°C. Answer: The final temperature is approximately 10.24°C.