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Question 1

Match corresponding quantities.

 Energy Answer 1Choose...Capacity to do workEnergy lost by a sytem + energy gained by surroundings = 0hf where f is the frequency of the light and h is Planck's constant0.5 x mass x velocity squaredmgh where m = mass, g = acceleration due to gravity, and h = heigth Kinetic energy Answer 2Choose...Capacity to do workEnergy lost by a sytem + energy gained by surroundings = 0hf where f is the frequency of the light and h is Planck's constant0.5 x mass x velocity squaredmgh where m = mass, g = acceleration due to gravity, and h = heigth Photon energy Answer 3Choose...Capacity to do workEnergy lost by a sytem + energy gained by surroundings = 0hf where f is the frequency of the light and h is Planck's constant0.5 x mass x velocity squaredmgh where m = mass, g = acceleration due to gravity, and h = heigth Conservation of energy Answer 4Choose...Capacity to do workEnergy lost by a sytem + energy gained by surroundings = 0hf where f is the frequency of the light and h is Planck's constant0.5 x mass x velocity squaredmgh where m = mass, g = acceleration due to gravity, and h = heigth Potential energy Answer 5Choose...Capacity to do workEnergy lost by a sytem + energy gained by surroundings = 0hf where f is the frequency of the light and h is Planck's constant0.5 x mass x velocity squaredmgh where m = mass, g = acceleration due to gravity, and h = heigth Question 2

5400 J of heat are added to a system. The surroundings do 2300 J of work on system. What is ΔE in J? Explanation

ΔE = q + w

= 5400 J + 2300 J

=7700 J

Question 3

2200 J of heat are extracted from a system. The system does 1400 J of work on the surroundings. What is ΔE in J? Explanation

ΔE = q + w

= -2200 J -1400 J

= -3600 J

Question 4

The temperature of the system is fixed. 617 J of heat are added to the system. What is the work in J done on the system?

Help: Two equations for energy Explanation

Because the temperature is constant, ΔE = 0. So, ΔE = q + w = 0. So w = -q.

Question 5

The work done on a system for a certain adiabatic process is 73 J. What is ΔE in J? Explanation

Because the process is adiabatic, q=" 0." So, ΔE = q + w = w.

Question 6

How much energy in kJ is required to heat a calorimeter whose heat capacity is 567.5 J from 23.5oC to 55.7oC? Explanation

Don't forget to change to kJ.

q= CΔT

= (567.5 J) (55.7 - 23.5 oC) / 1000

=18.27

Question 7

What is the heat capacity of a calorimeter in J/oC if 352.1 J of heat raises its temperature by 7.9 oC? q=CΔT

352.1 = C (7.9o)

44.569 = C

Question 8

What is the heat capacity of 0.63 kg of water in J/oC? Explanation

Don't forget to change to g.

C = m Cs

= (630 g)(4.184)

C= 2,635.92

Question 9

The specific heat of iron is 0.45 J/(goC). How much heat in J is added to a 9.54 g iron nail to raise its temperature from 22oC to 569.4oC? Explanation

q=mCsΔT

q= (9.54 g)(0.45 )(569.4oC-22oC)

q=2,349.9882

Question 10

A bowl of 292 g of water is placed in a microwave oven that puts out 874 watts (J/s). How long would it take in seconds to increase the temperature of the water from 12.0oC to 55.0oC? Use SF. Explanation

q=mCsΔT

(292)(4.184)(43)

52,534.304 J/ (874 J/s)

60.107 s

Question 11

211.9 g of water is in a Styrofoam calorimeter of negligible heat capacity. The initial T of the water is 15.1oC. After 459.8 g of an unknown compound at 96.4oC is added, the equilibrium T is 26.9oC. What is the specific heat of the unknown compound in J/(goC)? Explanation

mCsΔT=mCsΔT

(211.9 g)(4.184)(26.9-15.1)= (459.8g)x(96.4-26.9)

0.327379=x

Question 12

A person takes a complicated switchback trail to the top of a mountain. For which of the following cases would the person end up with the same potential energy or height?
Select one or more:   Explanation

Aint about how fast I get there

Aint about what's waiting on the other side

It's the climb

Question 13

A process takes place at constant pressure. The volume changes and the temperature increases by 71.7oC. The heat capacity of the system at constant volume, CV, is 218.6 kJ/oC. What is ΔE in kJ? (Careful--pay attention to the units of the heat capacity.) ΔE = CvΔT

=(218.6)(71.7)

=15,673.62

Question 14

For which of the following is the enthalpy of formation, ΔHfo = 0?

Select one or more:    Explanation

Elements DO NOT have heat formation value.

Question 15

For which of the following reactions is the enthalpy change in the reaction, ΔHrxno, equal to ΔHfo for NH4OH(s)?
Select one: Explanation

An equation for the enthalpy of formation must only create one mol of product and have all substances in their standard states.

Question 16

Calculate the enthalpy of reaction for the following reaction:

2Fe2O3(s) + 3C(s) → 4Fe(s) + 3CO2(g)

ΔHfo(Fe2O3(s)) = -824.2 kJ/mol
ΔHfo(C(s)) = ?
ΔHfo(Fe(s)) = ?
ΔHfo(CO2(g)) = -393.5 kJ/mol Question 17

The previous reaction is
Select one: Explanation

ΔH= (Sum)products - (Sum)reactants

3(-393.5 kJ/mol) - 2 (-824.2 kJ/mol)

-1,180.5 + 1648.4

467.9 kJ/mol

Explanation

It is positive=" Endothermic

Question 18

Calculate the enthalpy of reaction for the following reaction:

Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g)

ΔHfo(Fe2O3(s)) = -824.2 kJ/mol
ΔHfo(CO(g)) = -110.5
ΔHfo(Fe(s)) = ?
ΔHfo(CO2(g)) = -393.5 kJ/mol

Use SF. Explanation

3(-393.5 kJ/mol) - (3(-110.5) + (-824.2))

-1,180.5 - (-331.5-824.2)

-1180.5 + 1155.7

-24.8 kJ/mol

Question 19

The previous reaction is
Select one: Explanation

It is negative = exothermic

Question 20

4900 J of heat are added to a system. It does 4900 J of work on the surroundings. What is ΔE in J? Explanation

ΔE = q + w

q= 4900J

w = -4900J (work on the surrounding or by the system)

ΔE = 4900 - 4900 = 0

Question 21

A gas is compressed from 67.3 L to 3.1 L at 0.60 atmospheres? In addition a total of 3900 J of heat are gained from warmer surroundings. What is the change in internal energy in kJ? Explanation

Compression mean Work is done on thesystem, hence +ve

W= -PΔV
W= -P (Vf  -Vi)

ΔV= 3.1 L -67.3 L
= -64.2 L

W= -0.60 * -64.3
= 38.58 atm.L * 101 J/ atm = 3908J

=  3900J  + 3908J = 7808J = 7.80 kJ

Question 22

3300 J of heat are lost from a gas to cooler surroundings. At the same time the gas is compressed and the surroundings do 4100 J of work on the gas. What is ΔE for the gas in J? Explanation

ΔE = q + w

q= -3300J

w = 4100J (work on the system)

ΔE = -3300 + 4100 = 800 J