PHYSICAL CHEMISTRY-STUDENT SOLN.MAN.
PHYSICAL CHEMISTRY-STUDENT SOLN.MAN.
2nd Edition
ISBN: 9781285074788
Author: Ball
Publisher: CENGAGE L
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Chapter 3, Problem 3.34E

A 5.33 -g piece of Cu metal is heated to 99.7 ° C in boiling water, then dropped into a calorimeter containing 99.53 g of H 2 O at 22.6 ° C . The calorimeter is sealed to the outside environment, and temperature equalizes. C p [ Cu ( s ) ] = 0.385 J / g K , C p [ H 2 O ] = 4.18 J / g K . (a) Discuss the process that occurs inside the calorimeter in terms of the zeroth and first laws of thermodynamics. (b) What is the final temperature inside the system? (c) What is the entropy change of the Cu ( s ) ? (d)What is the entropy change of the H 2 O ( l ) ? (e) What is the total entropy change in the system? (f) Discuss the process that occurs inside the calorimeter in terms of the second law thermodynamics. Do you expect it to be spontaneous?

Expert Solution
Check Mark
Interpretation Introduction

(a)

Interpretation:

The process that occurs inside the calorimeter in terms of the zeroth and first laws of thermodynamics is to be described.

Concept introduction:

The zeroth law of thermodynamics states that if system A is in thermal equilibrium with system B and system B is in thermal equilibrium with system C, then system A and system C are also in thermal equilibrium with each other. The first law of thermodynamics states that the total energy of an isolated system remains unchanged.

Answer to Problem 3.34E

In terms of the zeroth law of thermodynamics, all the systems that are in contact with each other are in equilibrium. The systems present in the calorimeter are water, copper, and walls of the calorimeter. This complete system is in thermal equilibrium.

In terms of the first of law of thermodynamics, the energy released by copper at 99.7°C is equal to the heat accepted by water at 22.6°C.

Explanation of Solution

The whole system of calorimeter can be divided into three systems that are water, copper metal, and the walls of the calorimeter. All these systems are in contact with each other. According to the zeroth law of thermodynamics, these systems must be in thermal equilibrium with each other.

The temperature of the water is 22.6°C.

The temperature of copper metal is 99.7°C.

The temperature of copper metal is higher than that of water. Therefore, to attain a thermal equilibrium, some amount of energy is lost by the copper metal and some amount of energy is gained by the water. According to the first law of thermodynamics, the energy released by copper is equal to the heat accepted by water.

Conclusion

In terms of the zeroth law of thermodynamics, all the systems that are in contact with each other are in equilibrium. The systems present in the calorimeter are water, copper, and the walls of the calorimeter. This complete system is in thermal equilibrium.

In terms of the first of law of thermodynamics, the energy released by copper at 99.7°C is equal to the heat accepted by water at 22.6°C.

Expert Solution
Check Mark
Interpretation Introduction

(b)

Interpretation:

The final temperature inside the calorimeter system is to be calculated.

Concept introduction:

The first law of thermodynamics states that the total energy of an isolated system remains unchanged. The specific heat capacity of a substance is defined as the amount of heat required to raise the temperature of one gram of that substance by unity. The specific heat capacity of a substance is represented by the formula as shown below.

c=qm(TfTi)

Answer to Problem 3.34E

The final temperature inside the calorimeter system is 295.978K.

Explanation of Solution

The initial temperature of the water is 22.6°C.

The initial temperature of water in Kelvin is shown below.

T=(273+22.6°C)K=295.6K

The initial temperature of copper metal is 99.7°C.

The initial temperature of copper metal in Kelvin is shown below.

T=(273+99.7°C)K=372.7K

The final temperature of both copper metal and water is the same and assumed to be Tf.

The mass of the copper metal is 5.33g.

The mass of water is 99.53g.

The specific heat capacity of water is 4.18J/gK.

The specific heat capacity of copper metal is 0.385J/gK.

The exchange of heat due to temperature change is represented by the expression as shown below.

q=mc(TfTi) …(1)

Where,

m represents the mass of the substance.

c represents the specific heat capacity of the substance.

Tf represents the final temperature of the substance.

Ti represents the initial temperature of the substance.

In terms of the first of law of thermodynamics, the energy released by copper at 99.7°C is equal to the heat accepted by water at 22.6°C. This is mathematically represented by the expression as shown below.

qw=qCu …(2)

Where,

qw represents the heat change of water.

qCu represents the heat change of copper.

Substitute equation (1) in equation (2).

mwcw(TfTi, w)=mCucCu(TfTi, Cu) …(3)

Rearrange the equation (3) for the value of Tf.

Tf=mCucCuTi, Cu+mwcwTi, wmCucCu+mwcw …(4)

Substitute the values of initial temperature, specific heat, and masses of water and copper metal in equation (4).

Tf=(5.33g)(0.385J/gK)(372.7K)+(99.53g)(4.18J/gK)(295.6K)(5.33g)(0.385J/gK)+(99.53g)(4.18J/gK)=(764.7990+122980.06422.05205+416.0354)K=295.978K

Therefore, the final temperature inside the calorimeter system is 295.978K.

Conclusion

The final temperature inside the calorimeter system is 295.978K.

Expert Solution
Check Mark
Interpretation Introduction

(c)

Interpretation:

The entropy change of the Cu(s) is to be calculated.

Concept introduction:

The term entropy is used to represent the randomness in a system. When a system moves from an ordered arrangement to a less ordered arrangement, then the entropy of the system increases. The second law of thermodynamics states that the entropy of the system either increases or remains the same.

Answer to Problem 3.34E

The entropy change of the Cu(s) is 0.473J/K.

Explanation of Solution

The initial temperature of copper metal is 99.7°C.

The initial temperature of copper metal in Kelvin is shown below.

T=(273+99.7°C)K=372.7K

The final temperature of copper is 295.978K.

The mass of the copper metal is 5.33g.

The specific heat capacity of copper metal is 0.385J/gK.

The entropy change for the temperature change is represented by the expression as shown below.

ΔS=mcln(TfTi) …(5)

Where,

m represents the mass of the substance.

c represents the specific heat capacity of the substance.

Tf represents the final temperature of the substance.

Ti represents the initial temperature of the substance.

Substitute the values of final temperature, initial temperature, specific heat, and mass of copper metal in equation (5).

ΔS=(5.33g)(0.385J/gK)ln(295.978K372.7K)=(5.33g)(0.385J/gK)(0.2305)=0.473J/K

Therefore, the entropy change of the Cu(s) is 0.473J/K.

Conclusion

The entropy change of the Cu(s) is 0.473J/K.

Expert Solution
Check Mark
Interpretation Introduction

(d)

Interpretation:

The entropy change of the H2O(l) is to be calculated.

Concept introduction:

The term entropy is used to represent the randomness in a system. When a system moves from an ordered arrangement to a less ordered arrangement, then the entropy of the system increases. The second law of thermodynamics states that the entropy of the system either increases or remains the same.

Answer to Problem 3.34E

The entropy change of the H2O(l) is 0.532J/K.

Explanation of Solution

The initial temperature of the water is 22.6°C.

The initial temperature of water in Kelvin is shown below.

T=(273+22.6°C)K=295.6K

The final temperature of the water is 295.978K.

The mass of the water is 99.53g.

The specific heat capacity of water is 4.18J/gK.

The entropy change for the temperature change is represented by the expression as shown below.

ΔS=mcln(TfTi) …(5)

Where,

m represents the mass of the substance.

c represents the specific heat capacity of the substance.

Tf represents the final temperature of the substance.

Ti represents the initial temperature of the substance.

Substitute the values of final temperature, initial temperature, specific heat, and mass of water in equation (5).

ΔS=(99.53g)(4.18J/gK)ln(295.978K295.6K)=(99.53g)(4.18J/gK)(1.2779×103)=0.532J/K

Therefore, the entropy change of the H2O(l) is 0.532J/K.

Conclusion

The entropy change of the H2O(l) is 0.532J/K.

Expert Solution
Check Mark
Interpretation Introduction

(e)

Interpretation:

The total entropy change in the calorimeter system is to be calculated.

Concept introduction:

The term entropy is used to represent the randomness in a system. When a system moves from an ordered arrangement to a less ordered arrangement, then the entropy of the system increases. The second law of thermodynamics states that the entropy of the system either increases or remains the same.

Answer to Problem 3.34E

The total entropy change in the calorimeter system is 0.059J/K.

Explanation of Solution

The entropy change of the H2O(l) is 0.532J/K.

The entropy change of the Cu(s) is 0.473J/K.

The total entropy change in the system is represented by the expression shown below.

ΔSt=ΔSW+ΔSCu

Where,

ΔSW represents the entropy change of water.

ΔSCu represents the entropy change of copper.

Substitute the value of ΔSW and ΔSCu in the above equation.

ΔSt=0.532J/K+(0.473J/K)=0.059J/K

Therefore, the total entropy change in the calorimeter system is 0.059J/K.

Conclusion

The total entropy change in the calorimeter system is 0.059J/K.

Expert Solution
Check Mark
Interpretation Introduction

(f)

Interpretation:

The process that occurs inside the calorimeter in terms of the second law thermodynamics is to be described. Whether the corresponding process is spontaneous or not is to be stated.

Concept introduction:

The term entropy is used to represent the randomness in a system. When a system moves from an ordered arrangement to a less ordered arrangement, then the entropy of the system increases. The second law of thermodynamics states that the entropy of the system either increases or remains the same.

Answer to Problem 3.34E

The process is spontaneous because the entropy change of the system is positive. According to the second law of thermodynamics, the entropy change of system must be positive for a spontaneous process.

Explanation of Solution

The total entropy change in the calorimeter system is 0.059J/K.

This value implies that the randomness in the system is increasing with time. The second law of thermodynamics states that the entropy of the system either increases or remains the same. The entropy change is positive that means the entropy of the system is increasing. Therefore, the process is spontaneous.

Conclusion

The process is spontaneous because the entropy change of the system is positive. According to the second law of thermodynamics, the entropy change of system must be positive for a spontaneous process.

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Chapter 3 Solutions

PHYSICAL CHEMISTRY-STUDENT SOLN.MAN.

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