Thermochemistry Unit Chapter 17
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) = J Water absorbed this amount of energy.
- q sample = q water q water = J q sample = J This amount of energy is released.
mcΔT = q sample [(92.0 g)( c )(25.2 o C – 55 o C)] = J c = J / 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)] = J ( g o C)( c ) = J
Problem #2: A coffee cup calorimeter contains 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 g sample of water absorb? How much heat did the 25.3 g sample of water release ? m 1 = 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) = J q 2 = J
mcΔT = q 2 [(25.3 g)(4.184 J/g o C)(32.4 o C – T i )] = J b) What is the initial temperature of the water that was added to the calorimeter? [ J/ o C ) (32.4 o C - T i )] = J 32.4 o C - T i = J / J/ o C 32.4 o C - T i = o C - T i = o C – 32.4 o C = o C T i = o C
Problem #3: A 2.4 g diamond is heated to o C and placed in a coffee cup calorimeter containing some water. The initial temperature of the water is o C, and the final temperature of the contents of the calorimeter is o C. The specific heat capacity of diamond is J/g o C. What mass of water was in the calorimeter? m diamond = 2.4 g T i = o C m water = ? T f = o C ΔT diamond = o C – o C = o C - q diamond = q water q diamond = (mcΔT) = (2.4 g)(0.519 J/g o C)(-59.1 o C) = J q water = J T f = o C
ΔT water = o C – o C = 0.70 o C q water = (mcΔT) m water = g J = ( m )(4.184 J/g o C)(0.70 o C) J = ( J/g)( m ) J / ( J/g) = m
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 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 J/g o C. a) What was the mass of the sample of iron? m iron = ? T i = 98 o C m water = 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 = o C
- q iron = q water = J q water = (mcΔT) = (125.2 g)(4.184 J/g o C)(2 o C) = J m iron = g q iron = (mcΔT) J =( m )(0.444 J/g o C)(-73.7 o C) J =( m )( J/g) J/ ( J/g) = m
Problem #4: b) You try…..