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Social Science4. A 15 g piece of iron [cp = 25.09 J/(mol * °C)] is heated to a temperature of 100°C and placed into a bucket containing 1.0 L of water [cp = 75.38 J/(mol * °C)] at 25°C. What is the final temperature of the piece of iron and the water.
4. A 15 g piece of iron [cp = 25.09 J/(mol * °C)] is heated to a temperature of 100°C and placed into a bucket containing 1.0 L of water [cp = 75.38 J/(mol * °C)] at 25°C. What is the final temperature of the piece of iron and the water.
Expert Answer
The concept behind this question is the heat released by the metal is absorbed by the water. This can be represented by an equation as given below.
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(4)aa()er (mxсxAT). - (m хсхдT).. Where metal water -(m xcx AT), metal water m = mass c specificheat capacity AT Change in temperature
The negative sign is indicating that the heat is released. The specific heat capacity is given in the unit of J/mol 0C, so mass should be converted into moles.
The calculation of moles are given below.
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15g mass Moles of Fe =0.2686mol Molar mass 55.845g mol 1000 mL 1g density x volume Mass of water = 1000 g 1000g mass Moles of water = 55.5092mol Molar mass 18.015g mol
Suppose, the final temperature for both iron and water is T. Then, by plugging all the val...
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(4)mea(q)w metal water -(mxcxAT) 0.2686mol x 25.09 J/ mol.°Cx (100°C- T) mxсxдT)_. metal water -55.5092 molx 75.38J / mol.° C x (25° C- T) 0.2686mol x 25.09 J / mol.°Cx (100°C- T)= 55.5092molx 75.38 J / mol° C x (T - 25°C) (100°C-T) T-25°C) 55.5092 mol x 75.38 J/mof C 0.2686 mol x 25.09 J/mol^C = 620.8896 100 C T (T-25°C)x 620.8896 = 620.8896 T -8022.24 °C 620.8896T T= 100°C +8022.24 °C 8122.24°C 621.8896T 8122.24° C = 8122.240 C Т- 621.8896 13.06° C
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