Thermochemistry I: Energy Transfer and Calorimetry
1. What amount of work (in J) is performed on the surroundings when a 1.0 L balloon at 745 mm Hg
at 25°C is heated to 45°C? (1 L.atm = 101.325 J)
w = − P ∆V
P = 745 mm Hg = 0.9803 atm
760 mmatmHg
Ti = 25°C + 273 K=298 K Tf = 45°C+273 K=318 K
Vf = Vi ( ) = 1.0 L (
Tf
Ti
318 K
298 K ) = 1.07 L
∆V = 1.07 L -1.0 L=0.07 L
w = −(0.9803 atm)(0.07 L) × 101.325 L⋅atm
J
= −7.0 J
2. What quantity of heat (in J) is necessary to raise 3.00 L of water (d=1.00 g/mL) from 22.0°C to
63.0°C?
q = mc∆T
1000 g
m = 3.00 L × = 3000 g ( ± 10 g)
1L
c = 4.184 g⋅J°C
∆T = 63.0°C − 22.0°C=41.0°C
q = (3000 g)(4.184 g⋅J°C )(41.0°C)=515,000 J
3. A 200.0 mL quantity of 0.40 M HCl was added to 200.0 mL of 0.40 M NaOH in a solution
(constant pressure) calorimeter. The temperature of each solution was 25.10°C before mixing.
After mixing the solution rose to a temperature of 26.60°C before beginning to cool. The heat
capacity of the calorimeter was determined by separate experiment to be 55 J/°C. What is ∆Hrxn
per mol of H2O formed? Assume the solutions have a density of 1.00 g/mL and their specific heats
are similar to water; c = 4.18 J/g.°C.
VHCl = 200.0 mL VNaOH = 200.0 mL
mHCl = 200.0 g mNaOH = 200.0 g
nHCl = 0.2000 L(0.40M)=0.080 mol HCl nNaOH = 0.080 mol NaOH
qrxn + qsoln + qcal = 0
qrxn = n∆H rxn qsoln = mC ∆T qcal = C ∆T
msoln = 400.0 g csoln = 4.184 J
g ⋅°C ∆T = 26.60°C − 25.10°C = 1.50°C
(0.080 mol)∆H + 400.0 g(4.184 g⋅J°C )(1.50°C) + 55 °CJ (1.50°C)=0
(0.080 mol)∆H = −2590.5 J
∆H = −32, 400 mol
J
1. What amount of work (in J) is performed on the surroundings when a 1.0 L balloon at 745 mm Hg
at 25°C is heated to 45°C? (1 L.atm = 101.325 J)
w = − P ∆V
P = 745 mm Hg = 0.9803 atm
760 mmatmHg
Ti = 25°C + 273 K=298 K Tf = 45°C+273 K=318 K
Vf = Vi ( ) = 1.0 L (
Tf
Ti
318 K
298 K ) = 1.07 L
∆V = 1.07 L -1.0 L=0.07 L
w = −(0.9803 atm)(0.07 L) × 101.325 L⋅atm
J
= −7.0 J
2. What quantity of heat (in J) is necessary to raise 3.00 L of water (d=1.00 g/mL) from 22.0°C to
63.0°C?
q = mc∆T
1000 g
m = 3.00 L × = 3000 g ( ± 10 g)
1L
c = 4.184 g⋅J°C
∆T = 63.0°C − 22.0°C=41.0°C
q = (3000 g)(4.184 g⋅J°C )(41.0°C)=515,000 J
3. A 200.0 mL quantity of 0.40 M HCl was added to 200.0 mL of 0.40 M NaOH in a solution
(constant pressure) calorimeter. The temperature of each solution was 25.10°C before mixing.
After mixing the solution rose to a temperature of 26.60°C before beginning to cool. The heat
capacity of the calorimeter was determined by separate experiment to be 55 J/°C. What is ∆Hrxn
per mol of H2O formed? Assume the solutions have a density of 1.00 g/mL and their specific heats
are similar to water; c = 4.18 J/g.°C.
VHCl = 200.0 mL VNaOH = 200.0 mL
mHCl = 200.0 g mNaOH = 200.0 g
nHCl = 0.2000 L(0.40M)=0.080 mol HCl nNaOH = 0.080 mol NaOH
qrxn + qsoln + qcal = 0
qrxn = n∆H rxn qsoln = mC ∆T qcal = C ∆T
msoln = 400.0 g csoln = 4.184 J
g ⋅°C ∆T = 26.60°C − 25.10°C = 1.50°C
(0.080 mol)∆H + 400.0 g(4.184 g⋅J°C )(1.50°C) + 55 °CJ (1.50°C)=0
(0.080 mol)∆H = −2590.5 J
∆H = −32, 400 mol
J
