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CHAPTER 17 BASIC HOMEWORK:

THERMODYNAMICS I

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Question 1

How much energy in kJ is required to heat a calorimeter whose heat capacity is 510.0 J/oC from 23.5oC to 53.0oC?

Correct

 

Explanation

 Don't forget to change to kJ.

Q= CΔ

=510 (52-23.5)=15045 J--> 15.045 kJ

 

Question 2

What is the heat capacity of a calorimeter in J/oC if 120.3 J of heat raises its temperature by 5.0 oC?

Correct

 

Explanation

 120.3 J = C(5)

C=24.06

 

Question 3

What is the heat capacity of 0.83 kg of water in J/oC? The specific heat of water is 4.184 J/(goC).

Specific heat definition and example

Correct

 

Explanation

 Q= mCΔ

=(830 g) (4.184) (1)

=3472.72 J/C

 

Question 4

The specific heat of iron is 0.45 J/(goC). How much heat in J is added to a 9.24 g iron nail to raise its temperature from 22oC to 478.6oC?

Specific heat 2nd example

Correct

 

Explanation

 Q= mCΔ

= (9.24)(0.45)(478.6-22)

=1898.5 J

 

Question 5

A bowl of 192 g of water is placed in a microwave oven that puts out 642 watts (J/s). How long would it take in seconds to increase the temperature of the water from 12.0oC to 55.2oC?

Help given in feedback.
Correct

 

Explanation

 1. Find how many J are needed.
2. Convert J to sec. For example, suppose the microwave puts out 500 watts that is 500 J/s. To convert J to s do you multiple by (500 J)/s or 1 s/(500 J)?


q=192(4.184)(55.2-12)

=34,703.7696 J

l = q/w = 34703.7696 J/642 J/s = 54.06 s

 

Question 6

285.5 g of water is in a Styrofoam calorimeter of negligible heat capacity. The initial T of the water is 21.4oC. After 276.0 g of an unknown compound at 75.2oC is added, the equilibrium T is 30.8oC. What is the specific heat of the unknown compound in J/(goC)?

Specific heat 3rd example

Correct

 

Explanation

 mCΔT =mCΔ

(276 g)(C) (30.8-75.2) = (285.5 g)(4.184)(30.8-21.4)

-12,254,4(C)=11228.6008

C=0.916

 

Question 7

What is the approximate specific heat in J/(goC) of a metal whose M.W. is 58.5 g/mol?

Law of Dulong et Petit

Correct

 

Explanation

 molar heat capacity of 25 J/mol

25J/mol C * mol/58.5 g = 0.42735

 

Question 8

Match corresponding quantities.

Energy defined and kinds of energy

Conservation of energy

Correct

Energy

Correct

Photon energy

Correct

Kinetic energy

Correct

Potential energy

Correct

 

 

 

Question 9

4000 J of heat are added to a system. The surroundings do 1600 J of work on system. What is ΔE in J?

Energy calculation

Correct

 

Explanation

ΔE= q + w

=4000 + 1600 

=5600

 

Question 10

6100 J of heat are extracted from a system. The system does 1300 J of work on the surroundings. What is ΔE in J?

Correct

 

Explanation

 ΔE= q + w

=(-6100)-1300

=-7400

 

Question 11



The temperature of the system is fixed. 274 J of heat are added to the system. What is the work in J done on the system?

Help given in feedback.
Correct

 

Explanation

 The following two equations are always true: ΔE = q + w and ΔE = CVΔT where CV is the heat capacity at constant volume. The second equation says that if the temperature of a system increases its internal energy, E, increases. What is ΔE if the temperature is constant?


 ΔE = CVΔT

ΔT=0 ;  ΔE =0

 ΔE = q+w

0=q+w

 

Question 12

The work done on a system for a certain adiabatic process is -255 J. What is ΔE in J?

Correct

 

 

 

Question 13

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?

Examples of state functions

Select one or more:
Correct Correct
Correct

 

Explanation

 All must be correct for credit.

 

Question 14

A process takes place at constant pressure. The volume changes and the temperature increases by 94.7oC. The heat capacity of the system at constant volume, CV, is 475.3 kJ/oC. What is ΔE in kJ?

Help given in feedback.
Correct

 

Explanation

 Because E is a state function, no matter how it is correctly calculated you will get the same answer. Thus, even though the volume is not constant, you can use ΔE = CVΔT.


ΔH= E + pΔV

=(q+w)-w

w=-pΔV

ΔH=q

ΔE=CΔT

= (475.3)(94.7)

=45,010.91

 

Question 15

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

Select one or more:

Correct


Correct


Correct

Correct


h. C (diamond state)

 

Explanation

graphite state is more stable at STP (standard pressure and temperature)

 

Question 16

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

The Enthalpy of formation of a compound is given by the reaction where all the reactants are elements in their standard state and the product is the desired compound.


Select one:




Correct

 

 

 

 

Question 17

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

 

Explanation

 products-reactants

3(-393.5)-2(-824.2)=467.9

 

Question 18

he previous reaction is

Select one:

Correct

 

 

 

Question 19

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

Correct

 

Explanation

 3(-393.5)-[(-824.2)+3(-110.5)] = -1180.5+1155.7 = -24.8

 

Question 20

The previous reaction is


Select one:

Correct