1. Thermochemistry Unit Chapter 17

2. Problem #1 (page 664): A 92.0 g sample of a substance, with a temperature of 55 o C, is placed in a large scale polystyrene calorimeter. The calorimeter contains 1.00 kg of water at 20.0 o C. The final temperature of the system is 25.2 o C. a) How much heat did the sample release? How much heat did the water absorb ? m sample = 92.0 g T i = 92.0 o C m water = 1.00 kg = 1000 g T i = 20.0 o C T f system = 25.2 o C - q sample = q water q water = (mcΔT) = (1000 g)(4.184 J/g o C)(25.2 o C – 20.0 o C) = 21 756.8 J  Water absorbed this amount of energy.

3. - q sample = q water q water = 21 756.8 J  q sample = -21 756.8 J This amount of energy is released.

4. mcΔT = q sample [(92.0 g)( c )(25.2 o C – 55 o C)] = -21 756.8 J c = -21 756.8 J / -737.55 g o C c = 7.94 J/g o C b) What is the specific heat capacity of the substance? [(92.0 g)( c )(-29.8 o C)] = -21 756.8 J (-737.55 g o C)( c ) = -21 756.8 J

5. Problem #2: A coffee cup calorimeter contains 125.3 g of water at 20.3 o C. To this water 25.3 g of water is added. The water is mixed. The final temperature is 32.4 o C. a) How much heat did the 125.3 g sample of water absorb? How much heat did the 25.3 g sample of water release ? m 1 = 125.3 g T i = 20.3 o C m 2 = 25.3 g T f = 32.4 o C ΔT = 32.4 o C – 20.3 o C = 12.1 o C q 1 = - q 2 q 1 = (mcΔT) = (125.3 g)(4.184 J/g o C)(12.1) = 6 343.49 J  q 2 = -6 343.49 J

6. mcΔT = q 2 [(25.3 g)(4.184 J/g o C)(32.4 o C – T i )] = -6 343.49 J b) What is the initial temperature of the water that was added to the calorimeter? [105.86 J/ o C ) (32.4 o C - T i )] = -6 343.49 J 32.4 o C - T i = -6 343.49 J / 105.86 J/ o C 32.4 o C - T i = -59.92 o C - T i = -59.92 o C – 32.4 o C = -92.32 o C  T i = 92.32 o C

7. Problem #3: A 2.4 g diamond is heated to 85.30 o C and placed in a coffee cup calorimeter containing some water. The initial temperature of the water is 25.50 o C, and the final temperature of the contents of the calorimeter is 26.20 o C. The specific heat capacity of diamond is 0.519 J/g o C. What mass of water was in the calorimeter? m diamond = 2.4 g T i = 85.30 o C m water = ? T f = 26.20 o C ΔT diamond = 26.20 o C – 85.30 o C = -59.1 o C - q diamond = q water q diamond = (mcΔT) = (2.4 g)(0.519 J/g o C)(-59.1 o C) = -73.61 J  q water = 73.61 J T f = 26.20 o C

8. ΔT water = 26.20 o C – 25.50 o C = 0.70 o C q water = (mcΔT)  m water = 25.13 g 73.61 J = ( m )(4.184 J/g o C)(0.70 o C) 73.61 J = (2.9288 J/g)( m ) 73.61 J / (2.9288 J/g) = m

9. Problem #4: A sample of iron is heated to 98.0 o C in a hot water bath. The iron is then transferred to a coffee-cup calorimeter, which contains 125.2 g of water at 22.3 o C. The iron and water are allowed to come to thermal equilibrium. The final temperature of the contents of the calorimeter is 24.3 o C. The specific heat capacity of iron is 0.444 J/g o C. a) What was the mass of the sample of iron? m iron = ? T i = 98 o C m water = 125.2 g T f = 24.3 o C ΔT water = 24.3 o C – 22.3 o C = 2 o C T i = 22.3 o C T f = 24.3 o C ΔT iron = 24.3 o C – 98 o C = -73.7 o C

10. - q iron = q water = -1 047.67 J q water = (mcΔT) = (125.2 g)(4.184 J/g o C)(2 o C) = 1 047.67 J  m iron = 32.02 g q iron = (mcΔT) - 1047.67 J =( m )(0.444 J/g o C)(-73.7 o C) - 1047.67 J =( m )(-32.72 J/g) - 1047.67 J/ (-32.72 J/g) = m

11. Problem #4: b) You try…..