Hwk 12 Solutions - kapur(rk7259 Homework 12 Lyon(52380 This...

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kapur (rk7259) – Homework 12 – Lyon – (52380) 1 This print-out should have 41 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. This HW assignment is due Tuesday, May 5,by 11PM. 001 10.0 points Consider a reaction occuring in a calorimeter. The heat produced by this reaction is symbol- ized by 1. Δ H f 2. Δ H 3. + q correct 4. Δ H fus Explanation: Calorimeters measure internal energy changes. Δ E = q + w where q = heat en- ergy and w = work. 002 10.0 points A piece of a newly synthesized material of mass 25.0 g at 80.0 C is placed in a calorime- ter containing 100.0 g of water at 20.0 C. If the final temperature of the system is 24.0 C, what is the specific heat capacity of this ma- terial? 1. 1 . 19 J · g 1 · ( C) 1 correct 2. 7 . 46 J · g 1 · ( C) 1 3. 0 . 30 J · g 1 · ( C) 1 4. 4 . 76 J · g 1 · ( C) 1 5. 0 . 84 J · g 1 · ( C) 1 Explanation: Δ T m = (24 C - 80 C) = - 56 C Δ T w = (24 C - 20 C) = 4 C C s w = 4 . 184 J · g 1 · C 1 m w = 100 g m m = 25 g Heat energy will flow out of the hot metal in to the cool water. The final temperature of BOTH items will be 24.0 C. We can set up two expressions for heat flow using heat capacity and set them equal and opposite to each other in sign: - m m C s , m Δ T m = m w C s , w Δ T w C s , m = - m w C s , w Δ T w m m Δ T m C s , m = - (100 g)(4 C) (25 g)( - 56 C) × (4 . 184 J · g 1 · C 1 ) = 1 . 19543 J · g 1 · C 1 003 10.0 points A reaction known to release 2.00 kJ of heat takes place in a calorimeter containing 0.200 L of solution and the temperature rose by 4 . 46 C. When 100 mL of nitric acid and 100 mL of sodium hydroxide were mixed in the same calorimeter, the temperature rose by 2 . 01 C. What is the heat output for the neu- tralization reaction? 1. 0 . 816 kJ 2. 0 . 0168 kJ 3. 0 . 901 kJ correct 4. 17 . 9 kJ 5. 0 . 448 kJ Explanation: q absorbed = 2 kJ = 2000 J V 1 = 0 . 2 L Δ T 1 = 4 . 46 C V 2 = 200 mL = 0 . 2 L Δ T 2 = 2 . 01 C The first reaction enables us to calibrate the calorimeter by finding its heat capacity: We look at the second reaction and note that the SAME volume of solution is used, so we can use the data from the first reaction to find the heat capacity of the calorimeter including 0.2 L of solution. This makes the second calcula- tion a little faster. (The alternative is to find the heat capacity of just the calorimeter and then consider the heat absorbed by the water separately; necessary if the second reaction did not use the same volume of solution).
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kapur (rk7259) – Homework 12 – Lyon – (52380) 2 For the first reaction (calorimeter + solu- tion): q absorbed = C Δ T 1 C = q absorbed Δ T 1 = 2000 J 4 . 46 C = 448 . 43 J / C . We use this in the second reaction: q = c Δ T 2 = (448 . 43 J / C) (2 . 01 C) = 901 . 345 J = 0 . 901345 kJ Since this is the heat absorbed by the calorimeter, then it was given out by the re- action.
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