Pre-Lab Question #1: A 140 gram piece of iron initially at 100°C is placed into 1kg of water initially at 25°C. What is the final equilibrium temperature? Pre-Lab Question #2: A 200 gram piece of "blue-metal-stuff" at 100°C is placed into 1500 grams of water initially at 25°C. If the final temperature of the two is 53°C, what is the heat capacity (specific heat) of the "blue-metal-stuff"? Pre-Lab Question #3: A quantity of ice (200grams at -5°C) is placed in 300grams of water at 25°C. What is the final temperature of the mixture? What is the mixture of ice and water? (HINT: calculating the final temperature is a little different in this example. It should be consistent with your answer to the latter question. Also note that ice and liquid water have different properties; both are in the table above.)
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For iron, the specific heat capacity is 0.45 J/g°C, and for water, it is 4.18 J/g°C. The heat lost by the iron will be equal to the heat gained by the water. So, we have: (140g)(0.45 J/g°C)(Tf - 100°C) = (1000g)(4.18 J/g°C)(25°C - Tf) Solving this equation for Show more…
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