Thermochemistry Unit Thermochemistry Unit Chapter 17Chapter 17

Problem #1 (page 664):A 92.0 g sample of a substance, with a temperature of 55oC, is placed in a large scale polystyrene calorimeter. The calorimeter contains 1.00 kg of water at 20.0oC. The final temperature of the system is 25.2oC.a) How much heat did the sample release? How much heat did the water absorb?

msample = 92.0 g

Ti = 92.0oC

mwater = 1.00 kg = 1000 g

Ti = 20.0oCTf system = 25.2oC

- qsample = qwater

qwater = (mcΔT) = (1000 g)(4.184 J/goC)(25.2oC – 20.0oC)

= 21 756.8 J Water absorbed this amount of energy.

- qsample = qwater

qwater = 21 756.8 J

qsample = -21 756.8 J

This amount of energy is released.

mcΔT = qsample

[(92.0 g)(c)(25.2oC – 55oC)] = -21 756.8 J

c = -21 756.8 J / -737.55 g • oC

c = 7.94 J/g • oC

b) What is the specific heat capacity of the substance?

[(92.0 g)(c)(-29.8oC)] = -21 756.8 J

(-737.55 g • oC)(c) = -21 756.8 J

Problem #2:A coffee cup calorimeter contains 125.3 g of water at 20.3oC. To this water 25.3 g of water is added. The water is mixed. The final temperature is 32.4oC.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?

m1 = 125.3 g

Ti = 20.3oC

m2 = 25.3 g

Tf = 32.4oC

ΔT = 32.4oC – 20.3oC = 12.1oC

q1 = - q2

q1 = (mcΔT) = (125.3 g)(4.184 J/goC)(12.1) = 6 343.49 J

q2 = -6 343.49 J

mcΔT = q2

[(25.3 g)(4.184 J/goC)(32.4oC – Ti)] = -6 343.49 J

b) What is the initial temperature of the water that was added to the calorimeter?

[105.86 J/oC)(32.4oC - Ti)] = -6 343.49 J

32.4oC - Ti= -6 343.49 J / 105.86 J/oC

32.4oC - Ti= -59.92oC

32.4oC - Ti= -59.92oC

- Ti= -59.92oC – 32.4oC = -92.32oC

Ti= 92.32oC

Problem #3:A 2.4 g diamond is heated to 85.30oC and placed in a coffee cup calorimeter containing some water. The initial temperature of the water is 25.50oC, and the final temperature of the contents of the calorimeter is 26.20oC. The specific heat capacity of diamond is 0.519 J/goC. What mass of water was in the calorimeter?

mdiamond = 2.4 g

Ti = 85.30oC

mwater = ?

Tf = 26.20oC

ΔTdiamond = 26.20oC – 85.30oC = -59.1oC

- qdiamond = qwater

qdiamond = (mcΔT) = (2.4 g)(0.519 J/goC)(-59.1oC) = -73.61 J

qwater= 73.61 J

Tf = 26.20oC

ΔTwater = 26.20oC – 25.50oC = 0.70oC

qwater = (mcΔT)

mwater= 25.13 g

73.61 J = (m)(4.184 J/goC)(0.70oC)

73.61 J = (2.9288 J/g)(m)

73.61 J / (2.9288 J/g) = m

Problem #4:A sample of iron is heated to 98.0oC in a hot water bath. The iron is then transferred to a coffee-cup calorimeter, which contains 125.2 g of water at 22.3oC. The iron and water are allowed to come to thermal equilibrium. The final temperature of the contents of the calorimeter is 24.3oC. The specific heat capacity of iron is 0.444 J/goC. a) What was the mass of the sample of iron?

miron = ?

Ti = 98oC

mwater = 125.2 g

Tf = 24.3oC

ΔTwater = 24.3oC – 22.3oC = 2oC

Ti= 22.3oC

Tf= 24.3oC

ΔTiron = 24.3oC – 98oC = -73.7oC

- qiron = qwater = -1 047.67 J

qwater = (mcΔT) = (125.2 g)(4.184 J/goC)(2oC) = 1 047.67 J

miron= 32.02 g

qiron = (mcΔT)

- 1047.67 J =(m)(0.444 J/goC)(-73.7oC)

- 1047.67 J =(m)(-32.72 J/g)

- 1047.67 J/ (-32.72 J/g) = m

Problem #4:b) You try…..