## CHAPTER 6 BASIC HOMEWORK

**Resources: **

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**Question 1**

Match corresponding quantities. Help: Definition of energy EnergyChoose…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 energyChoose…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 energyChoose…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 energyChoose…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 energyChoose…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? Help: Internal energy

**Explanation**

**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**

**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 Additional hint given in feedback

**Explanation**

**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? Help: Adiabatic process Additional hint given in feedback

**Explanation**

**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? Help: Calorimeter and heat capacity

**Explanation**

**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?

**Explanation**

**Explanation**

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? Help: Specific heat

**Explanation**

**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? Help: Specific heat example

**Explanation**

**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. Help: Heating time

**Explanation**

**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)? Help: Specific heat calculation

**Explanation**

**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? Help: State functions Select one or more:

**Explanation**

**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.) Help: Energy is state function

**Explanation**

**Explanation**

ΔE = CvΔT =(218.6)(71.7) =15,673.62

**Question 14**

For which of the following is the enthalpy of formation, ΔHfo = 0? Help: Enthalpy of elements Select one or more:

**Explanation**

**Explanation**

Remember compounds are not elements.

**Question 15**

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

**Explanation**

**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

Help: Enthalpy of reaction example

**Explanation**

**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 It is positive= Endothermic

**Question 17**

The previous reaction is Help: Definition of endothermic and exothermic reactions Select one:

**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**

**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 It is negative = exothermic

**Question 19**

The previous reaction isSelect one:

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