Five thermal energy transfers that occur.

Question

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Answer 1

Question

Chapter 11, Problem 30P

(a)

To determine

Five thermal energy transfers that occur.

(a)

Expert Solution

Explanation of Solution

Aluminum container at 20°C contains ethyl alcohol at 30°C and ice at 0°C . Heat transfer from hot to cold body occurs. The mixture and the container attains an equilibrium temperature.

Ice at −10°C increases in temperature and becomes Ice at 0°C , then phase change occurs from ice to cold water at 0°C , then the temperature of the cold water at 0°C raises to a value T. Temperature of the Aluminum container at 20°C raises to a value T and ethyl alcohol at 30°C temperature rises to a value. The mixture and the container attain an equilibrium temperature T.

Conclusion: All the five thermal energy transfer that occurs is identified.

(b)

To determine

Construct a table similar to table 11.5

(b)

Expert Solution

Explanation of Solution

A comprehensive table is constructed for all the five thermal energy transfers.

Q	m(kg)	c(J/kg⋅°C)	L(J/kg)	Tf(°C)	Ti(°C)	Expressions
Qice	1.00	2090	-	0	−10.0	micecice[Tf−Ti]=micecice[0−(−10.0°C)]
Qmelt	1.00	-	3.33×105	0	0	miceLf
Qwater	1.00	4186	-	T	0	micecwater[Tf−Ti]=micecwater[Tf−0]
QAl	0.500	900	-	T	20.0	mAlcAl[Tf−Ti]=mAlcAl[Tf−20.0°C]
QEAlc	6.00	2430	-	T	30.0	mEAlccEAlc[Tf−Ti]=mEAlccEAlc[Tf−30.0°C]

Conclusion:

The comprehensive table similar to table 11.5 is constructed.

(c)

To determine

Conservation of energy equation.

(c)

Expert Solution

Explanation of Solution

According to the conservation of energy, heat lost by one in a mixture is equal to heat gained by the other.

Summing up all the energy expressions and equating it to zero.

{micecice[0−(−10.0°C)]+miceLf+micecwater[Tf−0]+mAlcAl[Tf−20.0°C]+mEAlccEAlc[Tf−30.0°C]=0}

Conclusion:

The equation denoting the conservation of energy after summing up all the expression and setting it to zero is constructed.

(d)

To determine

The final equilibrium temperature.

(d)

Expert Solution

Answer to Problem 30P

The final equilibrium temperature is T=4.81°C

Explanation of Solution

Given Info: Mass of aluminum container is 0.500 kg, initial temperature of the aluminum container is 20.0°C , mass of ethyl alcohol is 6.00 kg, initial temperature of ethyl alcohol is 30.0°C , mass of ice is 100 kg, initial temperature of ice is −10.0°C .

Formula to calculate the heat gained by cold water is,

{micecice[0−(−10.0°C)]+miceLf+micecwater[Tf−0]+mAlcAl[Tf−20.0°C]+mEAlccEAlc[Tf−30.0°C]=0}

Rearrange in terms of Tf .

{micecice(10.0°C)+miceLf+micecwaterTf+mAlcAlTf−mAlcAl(20.0°C)+mEAlccEAlcTf−mEAlccEAlc(30.0°C)}=0micecwaterTf+mAlcAlTf+mEAlccEAlcTf={−micecice(10.0°C)−miceLf+mAlcAl(20.0°C)+mEAlccEAlc(30.0°C)}Tf={−mice[cice(10.0°C)+Lf]+mAlcAl(20.0°C)+mEAlccEAlc(30.0°C)}micecwater+mAlcAl+mEAlccEAlc

Substitute all quantities in the second through sixth columns into the last column and sum to find T.

Tf={−(1.00 kg)[(2090 J/kg°C)(10.0°C)+(3.33×105 J/kg)]+[(0.500 kg)(900 J/kg°C)(20.0°C)]+[(6.00 kg)(2430 J/kg°C)(30.0°C)]}((1.00 kg)(4186 J/kg°C)+(0.500 kg)(900 J/kg°C)+[(6.0 kg)(2430 J/kg°C)])=9250019216=4.81°C

Conclusion:

Therefore, the final equilibrium temperature is 4.81°C .

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Answer 2

Question

Chapter 11, Problem 30P

(a)

To determine

Five thermal energy transfers that occur.

(a)

Expert Solution

Explanation of Solution

Aluminum container at 20°C contains ethyl alcohol at 30°C and ice at 0°C . Heat transfer from hot to cold body occurs. The mixture and the container attains an equilibrium temperature.

Ice at −10°C increases in temperature and becomes Ice at 0°C , then phase change occurs from ice to cold water at 0°C , then the temperature of the cold water at 0°C raises to a value T. Temperature of the Aluminum container at 20°C raises to a value T and ethyl alcohol at 30°C temperature rises to a value. The mixture and the container attain an equilibrium temperature T.

Conclusion: All the five thermal energy transfer that occurs is identified.

(b)

To determine

Construct a table similar to table 11.5

(b)

Expert Solution

Explanation of Solution

A comprehensive table is constructed for all the five thermal energy transfers.

Q	m(kg)	c(J/kg⋅°C)	L(J/kg)	Tf(°C)	Ti(°C)	Expressions
Qice	1.00	2090	-	0	−10.0	micecice[Tf−Ti]=micecice[0−(−10.0°C)]
Qmelt	1.00	-	3.33×105	0	0	miceLf
Qwater	1.00	4186	-	T	0	micecwater[Tf−Ti]=micecwater[Tf−0]
QAl	0.500	900	-	T	20.0	mAlcAl[Tf−Ti]=mAlcAl[Tf−20.0°C]
QEAlc	6.00	2430	-	T	30.0	mEAlccEAlc[Tf−Ti]=mEAlccEAlc[Tf−30.0°C]

Conclusion:

The comprehensive table similar to table 11.5 is constructed.

(c)

To determine

Conservation of energy equation.

(c)

Expert Solution

Explanation of Solution

According to the conservation of energy, heat lost by one in a mixture is equal to heat gained by the other.

Summing up all the energy expressions and equating it to zero.

{micecice[0−(−10.0°C)]+miceLf+micecwater[Tf−0]+mAlcAl[Tf−20.0°C]+mEAlccEAlc[Tf−30.0°C]=0}

Conclusion:

The equation denoting the conservation of energy after summing up all the expression and setting it to zero is constructed.

(d)

To determine

The final equilibrium temperature.

(d)

Expert Solution

Answer to Problem 30P

The final equilibrium temperature is T=4.81°C

Explanation of Solution

Given Info: Mass of aluminum container is 0.500 kg, initial temperature of the aluminum container is 20.0°C , mass of ethyl alcohol is 6.00 kg, initial temperature of ethyl alcohol is 30.0°C , mass of ice is 100 kg, initial temperature of ice is −10.0°C .

Formula to calculate the heat gained by cold water is,

{micecice[0−(−10.0°C)]+miceLf+micecwater[Tf−0]+mAlcAl[Tf−20.0°C]+mEAlccEAlc[Tf−30.0°C]=0}

Rearrange in terms of Tf .

{micecice(10.0°C)+miceLf+micecwaterTf+mAlcAlTf−mAlcAl(20.0°C)+mEAlccEAlcTf−mEAlccEAlc(30.0°C)}=0micecwaterTf+mAlcAlTf+mEAlccEAlcTf={−micecice(10.0°C)−miceLf+mAlcAl(20.0°C)+mEAlccEAlc(30.0°C)}Tf={−mice[cice(10.0°C)+Lf]+mAlcAl(20.0°C)+mEAlccEAlc(30.0°C)}micecwater+mAlcAl+mEAlccEAlc

Substitute all quantities in the second through sixth columns into the last column and sum to find T.

Tf={−(1.00 kg)[(2090 J/kg°C)(10.0°C)+(3.33×105 J/kg)]+[(0.500 kg)(900 J/kg°C)(20.0°C)]+[(6.00 kg)(2430 J/kg°C)(30.0°C)]}((1.00 kg)(4186 J/kg°C)+(0.500 kg)(900 J/kg°C)+[(6.0 kg)(2430 J/kg°C)])=9250019216=4.81°C

Conclusion:

Therefore, the final equilibrium temperature is 4.81°C .

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Answer 3

Question

Chapter 11, Problem 30P

(a)

To determine

Five thermal energy transfers that occur.

(a)

Expert Solution

Explanation of Solution

Aluminum container at 20°C contains ethyl alcohol at 30°C and ice at 0°C . Heat transfer from hot to cold body occurs. The mixture and the container attains an equilibrium temperature.

Ice at −10°C increases in temperature and becomes Ice at 0°C , then phase change occurs from ice to cold water at 0°C , then the temperature of the cold water at 0°C raises to a value T. Temperature of the Aluminum container at 20°C raises to a value T and ethyl alcohol at 30°C temperature rises to a value. The mixture and the container attain an equilibrium temperature T.

Conclusion: All the five thermal energy transfer that occurs is identified.

(b)

To determine

Construct a table similar to table 11.5

(b)

Expert Solution

Explanation of Solution

A comprehensive table is constructed for all the five thermal energy transfers.

Q	m(kg)	c(J/kg⋅°C)	L(J/kg)	Tf(°C)	Ti(°C)	Expressions
Qice	1.00	2090	-	0	−10.0	micecice[Tf−Ti]=micecice[0−(−10.0°C)]
Qmelt	1.00	-	3.33×105	0	0	miceLf
Qwater	1.00	4186	-	T	0	micecwater[Tf−Ti]=micecwater[Tf−0]
QAl	0.500	900	-	T	20.0	mAlcAl[Tf−Ti]=mAlcAl[Tf−20.0°C]
QEAlc	6.00	2430	-	T	30.0	mEAlccEAlc[Tf−Ti]=mEAlccEAlc[Tf−30.0°C]

Conclusion:

The comprehensive table similar to table 11.5 is constructed.

(c)

To determine

Conservation of energy equation.

(c)

Expert Solution

Explanation of Solution

According to the conservation of energy, heat lost by one in a mixture is equal to heat gained by the other.

Summing up all the energy expressions and equating it to zero.

{micecice[0−(−10.0°C)]+miceLf+micecwater[Tf−0]+mAlcAl[Tf−20.0°C]+mEAlccEAlc[Tf−30.0°C]=0}

Conclusion:

The equation denoting the conservation of energy after summing up all the expression and setting it to zero is constructed.

(d)

To determine

The final equilibrium temperature.

(d)

Expert Solution

Answer to Problem 30P

The final equilibrium temperature is T=4.81°C

Explanation of Solution

Given Info: Mass of aluminum container is 0.500 kg, initial temperature of the aluminum container is 20.0°C , mass of ethyl alcohol is 6.00 kg, initial temperature of ethyl alcohol is 30.0°C , mass of ice is 100 kg, initial temperature of ice is −10.0°C .

Formula to calculate the heat gained by cold water is,

{micecice[0−(−10.0°C)]+miceLf+micecwater[Tf−0]+mAlcAl[Tf−20.0°C]+mEAlccEAlc[Tf−30.0°C]=0}

Rearrange in terms of Tf .

{micecice(10.0°C)+miceLf+micecwaterTf+mAlcAlTf−mAlcAl(20.0°C)+mEAlccEAlcTf−mEAlccEAlc(30.0°C)}=0micecwaterTf+mAlcAlTf+mEAlccEAlcTf={−micecice(10.0°C)−miceLf+mAlcAl(20.0°C)+mEAlccEAlc(30.0°C)}Tf={−mice[cice(10.0°C)+Lf]+mAlcAl(20.0°C)+mEAlccEAlc(30.0°C)}micecwater+mAlcAl+mEAlccEAlc

Substitute all quantities in the second through sixth columns into the last column and sum to find T.

Tf={−(1.00 kg)[(2090 J/kg°C)(10.0°C)+(3.33×105 J/kg)]+[(0.500 kg)(900 J/kg°C)(20.0°C)]+[(6.00 kg)(2430 J/kg°C)(30.0°C)]}((1.00 kg)(4186 J/kg°C)+(0.500 kg)(900 J/kg°C)+[(6.0 kg)(2430 J/kg°C)])=9250019216=4.81°C

Conclusion:

Therefore, the final equilibrium temperature is 4.81°C .

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Answer 4

Question

Chapter 11, Problem 30P

(a)

To determine

Five thermal energy transfers that occur.

(a)

Expert Solution

Explanation of Solution

Aluminum container at 20°C contains ethyl alcohol at 30°C and ice at 0°C . Heat transfer from hot to cold body occurs. The mixture and the container attains an equilibrium temperature.

Ice at −10°C increases in temperature and becomes Ice at 0°C , then phase change occurs from ice to cold water at 0°C , then the temperature of the cold water at 0°C raises to a value T. Temperature of the Aluminum container at 20°C raises to a value T and ethyl alcohol at 30°C temperature rises to a value. The mixture and the container attain an equilibrium temperature T.

Conclusion: All the five thermal energy transfer that occurs is identified.

(b)

To determine

Construct a table similar to table 11.5

(b)

Expert Solution

Explanation of Solution

A comprehensive table is constructed for all the five thermal energy transfers.

Q	m(kg)	c(J/kg⋅°C)	L(J/kg)	Tf(°C)	Ti(°C)	Expressions
Qice	1.00	2090	-	0	−10.0	micecice[Tf−Ti]=micecice[0−(−10.0°C)]
Qmelt	1.00	-	3.33×105	0	0	miceLf
Qwater	1.00	4186	-	T	0	micecwater[Tf−Ti]=micecwater[Tf−0]
QAl	0.500	900	-	T	20.0	mAlcAl[Tf−Ti]=mAlcAl[Tf−20.0°C]
QEAlc	6.00	2430	-	T	30.0	mEAlccEAlc[Tf−Ti]=mEAlccEAlc[Tf−30.0°C]

Conclusion:

The comprehensive table similar to table 11.5 is constructed.

(c)

To determine

Conservation of energy equation.

(c)

Expert Solution

Explanation of Solution

According to the conservation of energy, heat lost by one in a mixture is equal to heat gained by the other.

Summing up all the energy expressions and equating it to zero.

{micecice[0−(−10.0°C)]+miceLf+micecwater[Tf−0]+mAlcAl[Tf−20.0°C]+mEAlccEAlc[Tf−30.0°C]=0}

Conclusion:

The equation denoting the conservation of energy after summing up all the expression and setting it to zero is constructed.

(d)

To determine

The final equilibrium temperature.

(d)

Expert Solution

Answer to Problem 30P

The final equilibrium temperature is T=4.81°C

Explanation of Solution

Given Info: Mass of aluminum container is 0.500 kg, initial temperature of the aluminum container is 20.0°C , mass of ethyl alcohol is 6.00 kg, initial temperature of ethyl alcohol is 30.0°C , mass of ice is 100 kg, initial temperature of ice is −10.0°C .

Formula to calculate the heat gained by cold water is,

{micecice[0−(−10.0°C)]+miceLf+micecwater[Tf−0]+mAlcAl[Tf−20.0°C]+mEAlccEAlc[Tf−30.0°C]=0}

Rearrange in terms of Tf .

{micecice(10.0°C)+miceLf+micecwaterTf+mAlcAlTf−mAlcAl(20.0°C)+mEAlccEAlcTf−mEAlccEAlc(30.0°C)}=0micecwaterTf+mAlcAlTf+mEAlccEAlcTf={−micecice(10.0°C)−miceLf+mAlcAl(20.0°C)+mEAlccEAlc(30.0°C)}Tf={−mice[cice(10.0°C)+Lf]+mAlcAl(20.0°C)+mEAlccEAlc(30.0°C)}micecwater+mAlcAl+mEAlccEAlc

Substitute all quantities in the second through sixth columns into the last column and sum to find T.

Tf={−(1.00 kg)[(2090 J/kg°C)(10.0°C)+(3.33×105 J/kg)]+[(0.500 kg)(900 J/kg°C)(20.0°C)]+[(6.00 kg)(2430 J/kg°C)(30.0°C)]}((1.00 kg)(4186 J/kg°C)+(0.500 kg)(900 J/kg°C)+[(6.0 kg)(2430 J/kg°C)])=9250019216=4.81°C

Conclusion:

Therefore, the final equilibrium temperature is 4.81°C .

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Answer 5

Chapter 11, Problem 30P

(a)

To determine

Five thermal energy transfers that occur.

(a)

Expert Solution

Explanation of Solution

Aluminum container at 20°C contains ethyl alcohol at 30°C and ice at 0°C . Heat transfer from hot to cold body occurs. The mixture and the container attains an equilibrium temperature.

Ice at −10°C increases in temperature and becomes Ice at 0°C , then phase change occurs from ice to cold water at 0°C , then the temperature of the cold water at 0°C raises to a value T. Temperature of the Aluminum container at 20°C raises to a value T and ethyl alcohol at 30°C temperature rises to a value. The mixture and the container attain an equilibrium temperature T.

Conclusion: All the five thermal energy transfer that occurs is identified.

(b)

To determine

Construct a table similar to table 11.5

(b)

Expert Solution

Explanation of Solution

A comprehensive table is constructed for all the five thermal energy transfers.

Q	m(kg)	c(J/kg⋅°C)	L(J/kg)	Tf(°C)	Ti(°C)	Expressions
Qice	1.00	2090	-	0	−10.0	micecice[Tf−Ti]=micecice[0−(−10.0°C)]
Qmelt	1.00	-	3.33×105	0	0	miceLf
Qwater	1.00	4186	-	T	0	micecwater[Tf−Ti]=micecwater[Tf−0]
QAl	0.500	900	-	T	20.0	mAlcAl[Tf−Ti]=mAlcAl[Tf−20.0°C]
QEAlc	6.00	2430	-	T	30.0	mEAlccEAlc[Tf−Ti]=mEAlccEAlc[Tf−30.0°C]

Conclusion:

The comprehensive table similar to table 11.5 is constructed.

(c)

To determine

Conservation of energy equation.

(c)

Expert Solution

Explanation of Solution

According to the conservation of energy, heat lost by one in a mixture is equal to heat gained by the other.

Summing up all the energy expressions and equating it to zero.

{micecice[0−(−10.0°C)]+miceLf+micecwater[Tf−0]+mAlcAl[Tf−20.0°C]+mEAlccEAlc[Tf−30.0°C]=0}

Conclusion:

The equation denoting the conservation of energy after summing up all the expression and setting it to zero is constructed.

(d)

To determine

The final equilibrium temperature.

(d)

Expert Solution

Answer to Problem 30P

The final equilibrium temperature is T=4.81°C

Explanation of Solution

Given Info: Mass of aluminum container is 0.500 kg, initial temperature of the aluminum container is 20.0°C , mass of ethyl alcohol is 6.00 kg, initial temperature of ethyl alcohol is 30.0°C , mass of ice is 100 kg, initial temperature of ice is −10.0°C .

Formula to calculate the heat gained by cold water is,

{micecice[0−(−10.0°C)]+miceLf+micecwater[Tf−0]+mAlcAl[Tf−20.0°C]+mEAlccEAlc[Tf−30.0°C]=0}

Rearrange in terms of Tf .

{micecice(10.0°C)+miceLf+micecwaterTf+mAlcAlTf−mAlcAl(20.0°C)+mEAlccEAlcTf−mEAlccEAlc(30.0°C)}=0micecwaterTf+mAlcAlTf+mEAlccEAlcTf={−micecice(10.0°C)−miceLf+mAlcAl(20.0°C)+mEAlccEAlc(30.0°C)}Tf={−mice[cice(10.0°C)+Lf]+mAlcAl(20.0°C)+mEAlccEAlc(30.0°C)}micecwater+mAlcAl+mEAlccEAlc

Substitute all quantities in the second through sixth columns into the last column and sum to find T.

Tf={−(1.00 kg)[(2090 J/kg°C)(10.0°C)+(3.33×105 J/kg)]+[(0.500 kg)(900 J/kg°C)(20.0°C)]+[(6.00 kg)(2430 J/kg°C)(30.0°C)]}((1.00 kg)(4186 J/kg°C)+(0.500 kg)(900 J/kg°C)+[(6.0 kg)(2430 J/kg°C)])=9250019216=4.81°C

Conclusion:

Therefore, the final equilibrium temperature is 4.81°C .

Answer 6

Chapter 11, Problem 30P

(a)

To determine

Five thermal energy transfers that occur.

(a)

Expert Solution

Explanation of Solution

Aluminum container at 20°C contains ethyl alcohol at 30°C and ice at 0°C . Heat transfer from hot to cold body occurs. The mixture and the container attains an equilibrium temperature.

Ice at −10°C increases in temperature and becomes Ice at 0°C , then phase change occurs from ice to cold water at 0°C , then the temperature of the cold water at 0°C raises to a value T. Temperature of the Aluminum container at 20°C raises to a value T and ethyl alcohol at 30°C temperature rises to a value. The mixture and the container attain an equilibrium temperature T.

Conclusion: All the five thermal energy transfer that occurs is identified.

Answer 7

Chapter 11, Problem 30P

(a)

To determine

Five thermal energy transfers that occur.

Answer 8

(a)

Expert Solution

Explanation of Solution

Aluminum container at 20°C contains ethyl alcohol at 30°C and ice at 0°C . Heat transfer from hot to cold body occurs. The mixture and the container attains an equilibrium temperature.

Ice at −10°C increases in temperature and becomes Ice at 0°C , then phase change occurs from ice to cold water at 0°C , then the temperature of the cold water at 0°C raises to a value T. Temperature of the Aluminum container at 20°C raises to a value T and ethyl alcohol at 30°C temperature rises to a value. The mixture and the container attain an equilibrium temperature T.

Conclusion: All the five thermal energy transfer that occurs is identified.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

(b)

To determine

Construct a table similar to table 11.5

(b)

Expert Solution

Explanation of Solution

A comprehensive table is constructed for all the five thermal energy transfers.

Q	m(kg)	c(J/kg⋅°C)	L(J/kg)	Tf(°C)	Ti(°C)	Expressions
Qice	1.00	2090	-	0	−10.0	micecice[Tf−Ti]=micecice[0−(−10.0°C)]
Qmelt	1.00	-	3.33×105	0	0	miceLf
Qwater	1.00	4186	-	T	0	micecwater[Tf−Ti]=micecwater[Tf−0]
QAl	0.500	900	-	T	20.0	mAlcAl[Tf−Ti]=mAlcAl[Tf−20.0°C]
QEAlc	6.00	2430	-	T	30.0	mEAlccEAlc[Tf−Ti]=mEAlccEAlc[Tf−30.0°C]

Conclusion:

The comprehensive table similar to table 11.5 is constructed.

Answer 13

(b)

To determine

Construct a table similar to table 11.5

Answer 14

(b)

Expert Solution

Explanation of Solution

A comprehensive table is constructed for all the five thermal energy transfers.

Q	m(kg)	c(J/kg⋅°C)	L(J/kg)	Tf(°C)	Ti(°C)	Expressions
Qice	1.00	2090	-	0	−10.0	micecice[Tf−Ti]=micecice[0−(−10.0°C)]
Qmelt	1.00	-	3.33×105	0	0	miceLf
Qwater	1.00	4186	-	T	0	micecwater[Tf−Ti]=micecwater[Tf−0]
QAl	0.500	900	-	T	20.0	mAlcAl[Tf−Ti]=mAlcAl[Tf−20.0°C]
QEAlc	6.00	2430	-	T	30.0	mEAlccEAlc[Tf−Ti]=mEAlccEAlc[Tf−30.0°C]

Conclusion:

The comprehensive table similar to table 11.5 is constructed.

Answer 15

(b)

Expert Solution

Answer 16

(b)

Expert Solution

Answer 17

Expert Solution

Answer 18

(c)

To determine

Conservation of energy equation.

(c)

Expert Solution

Explanation of Solution

According to the conservation of energy, heat lost by one in a mixture is equal to heat gained by the other.

Summing up all the energy expressions and equating it to zero.

{micecice[0−(−10.0°C)]+miceLf+micecwater[Tf−0]+mAlcAl[Tf−20.0°C]+mEAlccEAlc[Tf−30.0°C]=0}

Conclusion:

The equation denoting the conservation of energy after summing up all the expression and setting it to zero is constructed.

Answer 19

(c)

To determine

Conservation of energy equation.

Answer 20

(c)

Expert Solution

Explanation of Solution

According to the conservation of energy, heat lost by one in a mixture is equal to heat gained by the other.

Summing up all the energy expressions and equating it to zero.

{micecice[0−(−10.0°C)]+miceLf+micecwater[Tf−0]+mAlcAl[Tf−20.0°C]+mEAlccEAlc[Tf−30.0°C]=0}

Conclusion:

The equation denoting the conservation of energy after summing up all the expression and setting it to zero is constructed.

Answer 21

(c)

Expert Solution

Answer 22

(c)

Expert Solution

Answer 23

Expert Solution

Answer 24

(d)

To determine

The final equilibrium temperature.

(d)

Expert Solution

Answer to Problem 30P

The final equilibrium temperature is T=4.81°C

Explanation of Solution

Given Info: Mass of aluminum container is 0.500 kg, initial temperature of the aluminum container is 20.0°C , mass of ethyl alcohol is 6.00 kg, initial temperature of ethyl alcohol is 30.0°C , mass of ice is 100 kg, initial temperature of ice is −10.0°C .

Formula to calculate the heat gained by cold water is,

{micecice[0−(−10.0°C)]+miceLf+micecwater[Tf−0]+mAlcAl[Tf−20.0°C]+mEAlccEAlc[Tf−30.0°C]=0}

Rearrange in terms of Tf .

{micecice(10.0°C)+miceLf+micecwaterTf+mAlcAlTf−mAlcAl(20.0°C)+mEAlccEAlcTf−mEAlccEAlc(30.0°C)}=0micecwaterTf+mAlcAlTf+mEAlccEAlcTf={−micecice(10.0°C)−miceLf+mAlcAl(20.0°C)+mEAlccEAlc(30.0°C)}Tf={−mice[cice(10.0°C)+Lf]+mAlcAl(20.0°C)+mEAlccEAlc(30.0°C)}micecwater+mAlcAl+mEAlccEAlc

Substitute all quantities in the second through sixth columns into the last column and sum to find T.

Tf={−(1.00 kg)[(2090 J/kg°C)(10.0°C)+(3.33×105 J/kg)]+[(0.500 kg)(900 J/kg°C)(20.0°C)]+[(6.00 kg)(2430 J/kg°C)(30.0°C)]}((1.00 kg)(4186 J/kg°C)+(0.500 kg)(900 J/kg°C)+[(6.0 kg)(2430 J/kg°C)])=9250019216=4.81°C

Conclusion:

Therefore, the final equilibrium temperature is 4.81°C .

Answer 25

(d)

To determine

The final equilibrium temperature.

Answer 26

(d)

Expert Solution

Answer to Problem 30P

The final equilibrium temperature is T=4.81°C

Answer 27

(d)

Expert Solution

Answer 28

(d)

Expert Solution

Answer 29

Expert Solution

Answer 30

Explanation of Solution

Given Info: Mass of aluminum container is 0.500 kg, initial temperature of the aluminum container is 20.0°C , mass of ethyl alcohol is 6.00 kg, initial temperature of ethyl alcohol is 30.0°C , mass of ice is 100 kg, initial temperature of ice is −10.0°C .

Formula to calculate the heat gained by cold water is,

{micecice[0−(−10.0°C)]+miceLf+micecwater[Tf−0]+mAlcAl[Tf−20.0°C]+mEAlccEAlc[Tf−30.0°C]=0}

Rearrange in terms of Tf .

{micecice(10.0°C)+miceLf+micecwaterTf+mAlcAlTf−mAlcAl(20.0°C)+mEAlccEAlcTf−mEAlccEAlc(30.0°C)}=0micecwaterTf+mAlcAlTf+mEAlccEAlcTf={−micecice(10.0°C)−miceLf+mAlcAl(20.0°C)+mEAlccEAlc(30.0°C)}Tf={−mice[cice(10.0°C)+Lf]+mAlcAl(20.0°C)+mEAlccEAlc(30.0°C)}micecwater+mAlcAl+mEAlccEAlc

Substitute all quantities in the second through sixth columns into the last column and sum to find T.

Tf={−(1.00 kg)[(2090 J/kg°C)(10.0°C)+(3.33×105 J/kg)]+[(0.500 kg)(900 J/kg°C)(20.0°C)]+[(6.00 kg)(2430 J/kg°C)(30.0°C)]}((1.00 kg)(4186 J/kg°C)+(0.500 kg)(900 J/kg°C)+[(6.0 kg)(2430 J/kg°C)])=9250019216=4.81°C