The volume of turpentine overflows.

Question

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Answer 1

Question

Chapter 19, Problem 19.21P

(a)

To determine

The volume of turpentine overflows.

(a)

Expert Solution

Answer to Problem 19.21P

The volume of turpentine overflows is 99.4 mL .

Explanation of Solution

Given info: The initial volume of aluminum cylinder and turpentine is 2.000 L , the initial temperature is 20 °C , and the final temperature in the cylinder is 80 °C .

Formula to calculate the change in volume of aluminum cylinder is,

ΔVAl=VβAlΔT

Here,

βAl is the coefficient of volume expansion of aluminum.

ΔT is the temperature difference.

Formula to calculate the change in volume of turpentine is,

ΔVt=VβtΔT

Here,

βt is the coefficient of volume expansion of turpentine.

The relation between coefficient of volume expansion of aluminum and coefficient of linear expansion of aluminum is,

βAl=3αAl (2)

Here,

αAl is the coefficient of linear expansion of aluminum.

The coefficient of linear expansion of aluminum is 24×10−6 (°C)−1 .

Substitute 24×10−6 (°C)−1 for αAl in the above equation.

βAl=3(24×10−6 (°C)−1)=72×10−6 (°C)−1

Thus, the coefficient of volume expansion of aluminum is 72×10−6 (°C)−1 .

Write the expression for the temperature difference,

ΔT=Tf−Ti

Substitute 20 °C for Ti and 80 °C for Tf in the above equation.

ΔT=80 °C−20 °C=60 °C

Thus, the temperature difference is 60 °C .

Write the expression for overflow volume of the turpentine,

ΔV′=ΔVt−ΔVAl

Substitute VβAlΔT for ΔVAl and VβtΔT for ΔV in the above equation.

ΔV′=VβtΔT−VβAlΔT=VΔT(βt−βAl) (2)

The coefficient of volume expansion of turpentine is 9×10−4 (°C)−1 .

Substitute 72×10−6 (°C)−1 for βAl , 2.000 L for V , 60 °C for ΔT and 9×10−4 °C−1 for βt in the equation (2).

ΔV′=(2.000 L)(103 cm31 L)(9×10−4 (°C−1)−72×10−6 (°C−1))(60 °C)=99.4 cm3×1 mL1 cm3=99.4 mL

Conclusion:

Therefore, the volume of turpentine overflows is 99.4 mL .

(b)

To determine

The volume of turpentine remaining in the cylinder at 80 °C .

(b)

Expert Solution

Answer to Problem 19.21P

The volume of turpentine remaining in the cylinder at 80 °C is 2.01 L .

Explanation of Solution

Given info: The initial volume of aluminum cylinder and turpentine is 2.000 L , the initial temperature is 20 °C , and the final temperature in the cylinder is 80 °C .

Write the expression for the new volume of turpentine in the cylinder,

Vnew=V(1+βtΔT)

The coefficient of volume expansion of turpentine is 9×10−4 (°C)−1 and from the part (a) the temperature difference is 60 °C .

Substitute 2.000 L for V , 9×10−4 (°C)−1 for βt and 60 °C for ΔT in the above equation.

Vnew=(2.000 L)[1+(9×10−4 (°C)−1)(60 °C)]=2.01 L

Conclusion:

Therefore, the volume of turpentine remaining in the cylinder at 80 °C is 2.01 L .

(c)

To determine

The distance for the cylinder’s rim does the turpentine surface recede.

(c)

Expert Solution

Answer to Problem 19.21P

The distance for the cylinder’s rim does the turpentine surface recede is 0.998 cm .

Explanation of Solution

Given info: The initial volume of aluminum cylinder and turpentine is 2.000 L , the final temperature is 20 °C , and the initial temperature in the cylinder is 80 °C .

Write the expression for the volume of the turpentine in the cylinder after it cools back to 20 °C ,

V1=Vnew(1+βtΔT′) (3)

Write the expression for temperature difference,

ΔT′=T2−T1

Here,

T2 is the final temperature.

T1 is the initial temperature.

Substitute 20 °C for T2 and 80 °C for T1 in the above equation.

ΔT′=20 °C−80 °C=−60 °C

Thus, the temperature difference is −60 °C and from part (a) the volume of turpentine remaining in the cylinder at 80 °C is 2.1 L .

Substitute −60 °C for ΔT′ , 2.1 L for Vnew and 9×10−4 °C−1 for βt in the equation (3).

V1=(2.1 L)(103 cm31 L)[1+(9×10−4 °C−1)(−60 °C)]=1900.17 cm3

Write the expression for the percentage of cylinder that is empty at 20 °C ,

%V=V−V1V

Substitute 1900.17 cm3 for V1 and 2.000 L for V in the above equation.

%V=(2.0 L)(103 cm31 L)−1900.17 cm3(2.0 L)(103 cm31 L)=4.99%

The distance for the cylinder’s rim does the turpentine surface recede is,

4.99%(20.0 cm)=0.998 cm

Conclusion:

Therefore, the distance for the cylinder’s rim does the turpentine surface recede is 0.998 cm .

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Answer 2

Question

Chapter 19, Problem 19.21P

(a)

To determine

The volume of turpentine overflows.

(a)

Expert Solution

Answer to Problem 19.21P

The volume of turpentine overflows is 99.4 mL .

Explanation of Solution

Given info: The initial volume of aluminum cylinder and turpentine is 2.000 L , the initial temperature is 20 °C , and the final temperature in the cylinder is 80 °C .

Formula to calculate the change in volume of aluminum cylinder is,

ΔVAl=VβAlΔT

Here,

βAl is the coefficient of volume expansion of aluminum.

ΔT is the temperature difference.

Formula to calculate the change in volume of turpentine is,

ΔVt=VβtΔT

Here,

βt is the coefficient of volume expansion of turpentine.

The relation between coefficient of volume expansion of aluminum and coefficient of linear expansion of aluminum is,

βAl=3αAl (2)

Here,

αAl is the coefficient of linear expansion of aluminum.

The coefficient of linear expansion of aluminum is 24×10−6 (°C)−1 .

Substitute 24×10−6 (°C)−1 for αAl in the above equation.

βAl=3(24×10−6 (°C)−1)=72×10−6 (°C)−1

Thus, the coefficient of volume expansion of aluminum is 72×10−6 (°C)−1 .

Write the expression for the temperature difference,

ΔT=Tf−Ti

Substitute 20 °C for Ti and 80 °C for Tf in the above equation.

ΔT=80 °C−20 °C=60 °C

Thus, the temperature difference is 60 °C .

Write the expression for overflow volume of the turpentine,

ΔV′=ΔVt−ΔVAl

Substitute VβAlΔT for ΔVAl and VβtΔT for ΔV in the above equation.

ΔV′=VβtΔT−VβAlΔT=VΔT(βt−βAl) (2)

The coefficient of volume expansion of turpentine is 9×10−4 (°C)−1 .

Substitute 72×10−6 (°C)−1 for βAl , 2.000 L for V , 60 °C for ΔT and 9×10−4 °C−1 for βt in the equation (2).

ΔV′=(2.000 L)(103 cm31 L)(9×10−4 (°C−1)−72×10−6 (°C−1))(60 °C)=99.4 cm3×1 mL1 cm3=99.4 mL

Conclusion:

Therefore, the volume of turpentine overflows is 99.4 mL .

(b)

To determine

The volume of turpentine remaining in the cylinder at 80 °C .

(b)

Expert Solution

Answer to Problem 19.21P

The volume of turpentine remaining in the cylinder at 80 °C is 2.01 L .

Explanation of Solution

Given info: The initial volume of aluminum cylinder and turpentine is 2.000 L , the initial temperature is 20 °C , and the final temperature in the cylinder is 80 °C .

Write the expression for the new volume of turpentine in the cylinder,

Vnew=V(1+βtΔT)

The coefficient of volume expansion of turpentine is 9×10−4 (°C)−1 and from the part (a) the temperature difference is 60 °C .

Substitute 2.000 L for V , 9×10−4 (°C)−1 for βt and 60 °C for ΔT in the above equation.

Vnew=(2.000 L)[1+(9×10−4 (°C)−1)(60 °C)]=2.01 L

Conclusion:

Therefore, the volume of turpentine remaining in the cylinder at 80 °C is 2.01 L .

(c)

To determine

The distance for the cylinder’s rim does the turpentine surface recede.

(c)

Expert Solution

Answer to Problem 19.21P

The distance for the cylinder’s rim does the turpentine surface recede is 0.998 cm .

Explanation of Solution

Given info: The initial volume of aluminum cylinder and turpentine is 2.000 L , the final temperature is 20 °C , and the initial temperature in the cylinder is 80 °C .

Write the expression for the volume of the turpentine in the cylinder after it cools back to 20 °C ,

V1=Vnew(1+βtΔT′) (3)

Write the expression for temperature difference,

ΔT′=T2−T1

Here,

T2 is the final temperature.

T1 is the initial temperature.

Substitute 20 °C for T2 and 80 °C for T1 in the above equation.

ΔT′=20 °C−80 °C=−60 °C

Thus, the temperature difference is −60 °C and from part (a) the volume of turpentine remaining in the cylinder at 80 °C is 2.1 L .

Substitute −60 °C for ΔT′ , 2.1 L for Vnew and 9×10−4 °C−1 for βt in the equation (3).

V1=(2.1 L)(103 cm31 L)[1+(9×10−4 °C−1)(−60 °C)]=1900.17 cm3

Write the expression for the percentage of cylinder that is empty at 20 °C ,

%V=V−V1V

Substitute 1900.17 cm3 for V1 and 2.000 L for V in the above equation.

%V=(2.0 L)(103 cm31 L)−1900.17 cm3(2.0 L)(103 cm31 L)=4.99%

The distance for the cylinder’s rim does the turpentine surface recede is,

4.99%(20.0 cm)=0.998 cm

Conclusion:

Therefore, the distance for the cylinder’s rim does the turpentine surface recede is 0.998 cm .

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Answer 3

Question

Chapter 19, Problem 19.21P

(a)

To determine

The volume of turpentine overflows.

(a)

Expert Solution

Answer to Problem 19.21P

The volume of turpentine overflows is 99.4 mL .

Explanation of Solution

Given info: The initial volume of aluminum cylinder and turpentine is 2.000 L , the initial temperature is 20 °C , and the final temperature in the cylinder is 80 °C .

Formula to calculate the change in volume of aluminum cylinder is,

ΔVAl=VβAlΔT

Here,

βAl is the coefficient of volume expansion of aluminum.

ΔT is the temperature difference.

Formula to calculate the change in volume of turpentine is,

ΔVt=VβtΔT

Here,

βt is the coefficient of volume expansion of turpentine.

The relation between coefficient of volume expansion of aluminum and coefficient of linear expansion of aluminum is,

βAl=3αAl (2)

Here,

αAl is the coefficient of linear expansion of aluminum.

The coefficient of linear expansion of aluminum is 24×10−6 (°C)−1 .

Substitute 24×10−6 (°C)−1 for αAl in the above equation.

βAl=3(24×10−6 (°C)−1)=72×10−6 (°C)−1

Thus, the coefficient of volume expansion of aluminum is 72×10−6 (°C)−1 .

Write the expression for the temperature difference,

ΔT=Tf−Ti

Substitute 20 °C for Ti and 80 °C for Tf in the above equation.

ΔT=80 °C−20 °C=60 °C

Thus, the temperature difference is 60 °C .

Write the expression for overflow volume of the turpentine,

ΔV′=ΔVt−ΔVAl

Substitute VβAlΔT for ΔVAl and VβtΔT for ΔV in the above equation.

ΔV′=VβtΔT−VβAlΔT=VΔT(βt−βAl) (2)

The coefficient of volume expansion of turpentine is 9×10−4 (°C)−1 .

Substitute 72×10−6 (°C)−1 for βAl , 2.000 L for V , 60 °C for ΔT and 9×10−4 °C−1 for βt in the equation (2).

ΔV′=(2.000 L)(103 cm31 L)(9×10−4 (°C−1)−72×10−6 (°C−1))(60 °C)=99.4 cm3×1 mL1 cm3=99.4 mL

Conclusion:

Therefore, the volume of turpentine overflows is 99.4 mL .

(b)

To determine

The volume of turpentine remaining in the cylinder at 80 °C .

(b)

Expert Solution

Answer to Problem 19.21P

The volume of turpentine remaining in the cylinder at 80 °C is 2.01 L .

Explanation of Solution

Given info: The initial volume of aluminum cylinder and turpentine is 2.000 L , the initial temperature is 20 °C , and the final temperature in the cylinder is 80 °C .

Write the expression for the new volume of turpentine in the cylinder,

Vnew=V(1+βtΔT)

The coefficient of volume expansion of turpentine is 9×10−4 (°C)−1 and from the part (a) the temperature difference is 60 °C .

Substitute 2.000 L for V , 9×10−4 (°C)−1 for βt and 60 °C for ΔT in the above equation.

Vnew=(2.000 L)[1+(9×10−4 (°C)−1)(60 °C)]=2.01 L

Conclusion:

Therefore, the volume of turpentine remaining in the cylinder at 80 °C is 2.01 L .

(c)

To determine

The distance for the cylinder’s rim does the turpentine surface recede.

(c)

Expert Solution

Answer to Problem 19.21P

The distance for the cylinder’s rim does the turpentine surface recede is 0.998 cm .

Explanation of Solution

Given info: The initial volume of aluminum cylinder and turpentine is 2.000 L , the final temperature is 20 °C , and the initial temperature in the cylinder is 80 °C .

Write the expression for the volume of the turpentine in the cylinder after it cools back to 20 °C ,

V1=Vnew(1+βtΔT′) (3)

Write the expression for temperature difference,

ΔT′=T2−T1

Here,

T2 is the final temperature.

T1 is the initial temperature.

Substitute 20 °C for T2 and 80 °C for T1 in the above equation.

ΔT′=20 °C−80 °C=−60 °C

Thus, the temperature difference is −60 °C and from part (a) the volume of turpentine remaining in the cylinder at 80 °C is 2.1 L .

Substitute −60 °C for ΔT′ , 2.1 L for Vnew and 9×10−4 °C−1 for βt in the equation (3).

V1=(2.1 L)(103 cm31 L)[1+(9×10−4 °C−1)(−60 °C)]=1900.17 cm3

Write the expression for the percentage of cylinder that is empty at 20 °C ,

%V=V−V1V

Substitute 1900.17 cm3 for V1 and 2.000 L for V in the above equation.

%V=(2.0 L)(103 cm31 L)−1900.17 cm3(2.0 L)(103 cm31 L)=4.99%

The distance for the cylinder’s rim does the turpentine surface recede is,

4.99%(20.0 cm)=0.998 cm

Conclusion:

Therefore, the distance for the cylinder’s rim does the turpentine surface recede is 0.998 cm .

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Answer 4

Question

Chapter 19, Problem 19.21P

(a)

To determine

The volume of turpentine overflows.

(a)

Expert Solution

Answer to Problem 19.21P

The volume of turpentine overflows is 99.4 mL .

Explanation of Solution

Given info: The initial volume of aluminum cylinder and turpentine is 2.000 L , the initial temperature is 20 °C , and the final temperature in the cylinder is 80 °C .

Formula to calculate the change in volume of aluminum cylinder is,

ΔVAl=VβAlΔT

Here,

βAl is the coefficient of volume expansion of aluminum.

ΔT is the temperature difference.

Formula to calculate the change in volume of turpentine is,

ΔVt=VβtΔT

Here,

βt is the coefficient of volume expansion of turpentine.

The relation between coefficient of volume expansion of aluminum and coefficient of linear expansion of aluminum is,

βAl=3αAl (2)

Here,

αAl is the coefficient of linear expansion of aluminum.

The coefficient of linear expansion of aluminum is 24×10−6 (°C)−1 .

Substitute 24×10−6 (°C)−1 for αAl in the above equation.

βAl=3(24×10−6 (°C)−1)=72×10−6 (°C)−1

Thus, the coefficient of volume expansion of aluminum is 72×10−6 (°C)−1 .

Write the expression for the temperature difference,

ΔT=Tf−Ti

Substitute 20 °C for Ti and 80 °C for Tf in the above equation.

ΔT=80 °C−20 °C=60 °C

Thus, the temperature difference is 60 °C .

Write the expression for overflow volume of the turpentine,

ΔV′=ΔVt−ΔVAl

Substitute VβAlΔT for ΔVAl and VβtΔT for ΔV in the above equation.

ΔV′=VβtΔT−VβAlΔT=VΔT(βt−βAl) (2)

The coefficient of volume expansion of turpentine is 9×10−4 (°C)−1 .

Substitute 72×10−6 (°C)−1 for βAl , 2.000 L for V , 60 °C for ΔT and 9×10−4 °C−1 for βt in the equation (2).

ΔV′=(2.000 L)(103 cm31 L)(9×10−4 (°C−1)−72×10−6 (°C−1))(60 °C)=99.4 cm3×1 mL1 cm3=99.4 mL

Conclusion:

Therefore, the volume of turpentine overflows is 99.4 mL .

(b)

To determine

The volume of turpentine remaining in the cylinder at 80 °C .

(b)

Expert Solution

Answer to Problem 19.21P

The volume of turpentine remaining in the cylinder at 80 °C is 2.01 L .

Explanation of Solution

Given info: The initial volume of aluminum cylinder and turpentine is 2.000 L , the initial temperature is 20 °C , and the final temperature in the cylinder is 80 °C .

Write the expression for the new volume of turpentine in the cylinder,

Vnew=V(1+βtΔT)

The coefficient of volume expansion of turpentine is 9×10−4 (°C)−1 and from the part (a) the temperature difference is 60 °C .

Substitute 2.000 L for V , 9×10−4 (°C)−1 for βt and 60 °C for ΔT in the above equation.

Vnew=(2.000 L)[1+(9×10−4 (°C)−1)(60 °C)]=2.01 L

Conclusion:

Therefore, the volume of turpentine remaining in the cylinder at 80 °C is 2.01 L .

(c)

To determine

The distance for the cylinder’s rim does the turpentine surface recede.

(c)

Expert Solution

Answer to Problem 19.21P

The distance for the cylinder’s rim does the turpentine surface recede is 0.998 cm .

Explanation of Solution

Given info: The initial volume of aluminum cylinder and turpentine is 2.000 L , the final temperature is 20 °C , and the initial temperature in the cylinder is 80 °C .

Write the expression for the volume of the turpentine in the cylinder after it cools back to 20 °C ,

V1=Vnew(1+βtΔT′) (3)

Write the expression for temperature difference,

ΔT′=T2−T1

Here,

T2 is the final temperature.

T1 is the initial temperature.

Substitute 20 °C for T2 and 80 °C for T1 in the above equation.

ΔT′=20 °C−80 °C=−60 °C

Thus, the temperature difference is −60 °C and from part (a) the volume of turpentine remaining in the cylinder at 80 °C is 2.1 L .

Substitute −60 °C for ΔT′ , 2.1 L for Vnew and 9×10−4 °C−1 for βt in the equation (3).

V1=(2.1 L)(103 cm31 L)[1+(9×10−4 °C−1)(−60 °C)]=1900.17 cm3

Write the expression for the percentage of cylinder that is empty at 20 °C ,

%V=V−V1V

Substitute 1900.17 cm3 for V1 and 2.000 L for V in the above equation.

%V=(2.0 L)(103 cm31 L)−1900.17 cm3(2.0 L)(103 cm31 L)=4.99%

The distance for the cylinder’s rim does the turpentine surface recede is,

4.99%(20.0 cm)=0.998 cm

Conclusion:

Therefore, the distance for the cylinder’s rim does the turpentine surface recede is 0.998 cm .

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Answer 5

Chapter 19, Problem 19.21P

(a)

To determine

The volume of turpentine overflows.

(a)

Expert Solution

Answer to Problem 19.21P

The volume of turpentine overflows is 99.4 mL .

Explanation of Solution

Given info: The initial volume of aluminum cylinder and turpentine is 2.000 L , the initial temperature is 20 °C , and the final temperature in the cylinder is 80 °C .

Formula to calculate the change in volume of aluminum cylinder is,

ΔVAl=VβAlΔT

Here,

βAl is the coefficient of volume expansion of aluminum.

ΔT is the temperature difference.

Formula to calculate the change in volume of turpentine is,

ΔVt=VβtΔT

Here,

βt is the coefficient of volume expansion of turpentine.

The relation between coefficient of volume expansion of aluminum and coefficient of linear expansion of aluminum is,

βAl=3αAl (2)

Here,

αAl is the coefficient of linear expansion of aluminum.

The coefficient of linear expansion of aluminum is 24×10−6 (°C)−1 .

Substitute 24×10−6 (°C)−1 for αAl in the above equation.

βAl=3(24×10−6 (°C)−1)=72×10−6 (°C)−1

Thus, the coefficient of volume expansion of aluminum is 72×10−6 (°C)−1 .

Write the expression for the temperature difference,

ΔT=Tf−Ti

Substitute 20 °C for Ti and 80 °C for Tf in the above equation.

ΔT=80 °C−20 °C=60 °C

Thus, the temperature difference is 60 °C .

Write the expression for overflow volume of the turpentine,

ΔV′=ΔVt−ΔVAl

Substitute VβAlΔT for ΔVAl and VβtΔT for ΔV in the above equation.

ΔV′=VβtΔT−VβAlΔT=VΔT(βt−βAl) (2)

The coefficient of volume expansion of turpentine is 9×10−4 (°C)−1 .

Substitute 72×10−6 (°C)−1 for βAl , 2.000 L for V , 60 °C for ΔT and 9×10−4 °C−1 for βt in the equation (2).

ΔV′=(2.000 L)(103 cm31 L)(9×10−4 (°C−1)−72×10−6 (°C−1))(60 °C)=99.4 cm3×1 mL1 cm3=99.4 mL

Conclusion:

Therefore, the volume of turpentine overflows is 99.4 mL .

(b)

To determine

The volume of turpentine remaining in the cylinder at 80 °C .

(b)

Expert Solution

Answer to Problem 19.21P

The volume of turpentine remaining in the cylinder at 80 °C is 2.01 L .

Explanation of Solution

Given info: The initial volume of aluminum cylinder and turpentine is 2.000 L , the initial temperature is 20 °C , and the final temperature in the cylinder is 80 °C .

Write the expression for the new volume of turpentine in the cylinder,

Vnew=V(1+βtΔT)

The coefficient of volume expansion of turpentine is 9×10−4 (°C)−1 and from the part (a) the temperature difference is 60 °C .

Substitute 2.000 L for V , 9×10−4 (°C)−1 for βt and 60 °C for ΔT in the above equation.

Vnew=(2.000 L)[1+(9×10−4 (°C)−1)(60 °C)]=2.01 L

Conclusion:

Therefore, the volume of turpentine remaining in the cylinder at 80 °C is 2.01 L .

(c)

To determine

The distance for the cylinder’s rim does the turpentine surface recede.

(c)

Expert Solution

Answer to Problem 19.21P

The distance for the cylinder’s rim does the turpentine surface recede is 0.998 cm .

Explanation of Solution

Given info: The initial volume of aluminum cylinder and turpentine is 2.000 L , the final temperature is 20 °C , and the initial temperature in the cylinder is 80 °C .

Write the expression for the volume of the turpentine in the cylinder after it cools back to 20 °C ,

V1=Vnew(1+βtΔT′) (3)

Write the expression for temperature difference,

ΔT′=T2−T1

Here,

T2 is the final temperature.

T1 is the initial temperature.

Substitute 20 °C for T2 and 80 °C for T1 in the above equation.

ΔT′=20 °C−80 °C=−60 °C

Thus, the temperature difference is −60 °C and from part (a) the volume of turpentine remaining in the cylinder at 80 °C is 2.1 L .

Substitute −60 °C for ΔT′ , 2.1 L for Vnew and 9×10−4 °C−1 for βt in the equation (3).

V1=(2.1 L)(103 cm31 L)[1+(9×10−4 °C−1)(−60 °C)]=1900.17 cm3

Write the expression for the percentage of cylinder that is empty at 20 °C ,

%V=V−V1V

Substitute 1900.17 cm3 for V1 and 2.000 L for V in the above equation.

%V=(2.0 L)(103 cm31 L)−1900.17 cm3(2.0 L)(103 cm31 L)=4.99%

The distance for the cylinder’s rim does the turpentine surface recede is,

4.99%(20.0 cm)=0.998 cm

Conclusion:

Therefore, the distance for the cylinder’s rim does the turpentine surface recede is 0.998 cm .

Answer 6

Chapter 19, Problem 19.21P

(a)

To determine

The volume of turpentine overflows.

(a)

Expert Solution

Answer to Problem 19.21P

The volume of turpentine overflows is 99.4 mL .

Explanation of Solution

Given info: The initial volume of aluminum cylinder and turpentine is 2.000 L , the initial temperature is 20 °C , and the final temperature in the cylinder is 80 °C .

Formula to calculate the change in volume of aluminum cylinder is,

ΔVAl=VβAlΔT

Here,

βAl is the coefficient of volume expansion of aluminum.

ΔT is the temperature difference.

Formula to calculate the change in volume of turpentine is,

ΔVt=VβtΔT

Here,

βt is the coefficient of volume expansion of turpentine.

The relation between coefficient of volume expansion of aluminum and coefficient of linear expansion of aluminum is,

βAl=3αAl (2)

Here,

αAl is the coefficient of linear expansion of aluminum.

The coefficient of linear expansion of aluminum is 24×10−6 (°C)−1 .

Substitute 24×10−6 (°C)−1 for αAl in the above equation.

βAl=3(24×10−6 (°C)−1)=72×10−6 (°C)−1

Thus, the coefficient of volume expansion of aluminum is 72×10−6 (°C)−1 .

Write the expression for the temperature difference,

ΔT=Tf−Ti

Substitute 20 °C for Ti and 80 °C for Tf in the above equation.

ΔT=80 °C−20 °C=60 °C

Thus, the temperature difference is 60 °C .

Write the expression for overflow volume of the turpentine,

ΔV′=ΔVt−ΔVAl

Substitute VβAlΔT for ΔVAl and VβtΔT for ΔV in the above equation.

ΔV′=VβtΔT−VβAlΔT=VΔT(βt−βAl) (2)

The coefficient of volume expansion of turpentine is 9×10−4 (°C)−1 .

Substitute 72×10−6 (°C)−1 for βAl , 2.000 L for V , 60 °C for ΔT and 9×10−4 °C−1 for βt in the equation (2).

ΔV′=(2.000 L)(103 cm31 L)(9×10−4 (°C−1)−72×10−6 (°C−1))(60 °C)=99.4 cm3×1 mL1 cm3=99.4 mL

Conclusion:

Therefore, the volume of turpentine overflows is 99.4 mL .

Answer 7

Chapter 19, Problem 19.21P

(a)

To determine

The volume of turpentine overflows.

Answer 8

(a)

Expert Solution

Answer to Problem 19.21P

The volume of turpentine overflows is 99.4 mL .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given info: The initial volume of aluminum cylinder and turpentine is 2.000 L , the initial temperature is 20 °C , and the final temperature in the cylinder is 80 °C .

Formula to calculate the change in volume of aluminum cylinder is,

ΔVAl=VβAlΔT

Here,

βAl is the coefficient of volume expansion of aluminum.

ΔT is the temperature difference.

Formula to calculate the change in volume of turpentine is,

ΔVt=VβtΔT

Here,

βt is the coefficient of volume expansion of turpentine.

The relation between coefficient of volume expansion of aluminum and coefficient of linear expansion of aluminum is,

βAl=3αAl (2)

Here,

αAl is the coefficient of linear expansion of aluminum.

The coefficient of linear expansion of aluminum is 24×10−6 (°C)−1 .

Substitute 24×10−6 (°C)−1 for αAl in the above equation.

βAl=3(24×10−6 (°C)−1)=72×10−6 (°C)−1

Thus, the coefficient of volume expansion of aluminum is 72×10−6 (°C)−1 .

Write the expression for the temperature difference,

ΔT=Tf−Ti

Substitute 20 °C for Ti and 80 °C for Tf in the above equation.

ΔT=80 °C−20 °C=60 °C

Thus, the temperature difference is 60 °C .

Write the expression for overflow volume of the turpentine,

ΔV′=ΔVt−ΔVAl

Substitute VβAlΔT for ΔVAl and VβtΔT for ΔV in the above equation.

ΔV′=VβtΔT−VβAlΔT=VΔT(βt−βAl) (2)

The coefficient of volume expansion of turpentine is 9×10−4 (°C)−1 .

Substitute 72×10−6 (°C)−1 for βAl , 2.000 L for V , 60 °C for ΔT and 9×10−4 °C−1 for βt in the equation (2).

ΔV′=(2.000 L)(103 cm31 L)(9×10−4 (°C−1)−72×10−6 (°C−1))(60 °C)=99.4 cm3×1 mL1 cm3=99.4 mL

Conclusion:

Therefore, the volume of turpentine overflows is 99.4 mL .

Answer 13

(b)

To determine

The volume of turpentine remaining in the cylinder at 80 °C .

(b)

Expert Solution

Answer to Problem 19.21P

The volume of turpentine remaining in the cylinder at 80 °C is 2.01 L .

Explanation of Solution

Given info: The initial volume of aluminum cylinder and turpentine is 2.000 L , the initial temperature is 20 °C , and the final temperature in the cylinder is 80 °C .

Write the expression for the new volume of turpentine in the cylinder,

Vnew=V(1+βtΔT)

The coefficient of volume expansion of turpentine is 9×10−4 (°C)−1 and from the part (a) the temperature difference is 60 °C .

Substitute 2.000 L for V , 9×10−4 (°C)−1 for βt and 60 °C for ΔT in the above equation.

Vnew=(2.000 L)[1+(9×10−4 (°C)−1)(60 °C)]=2.01 L

Conclusion:

Therefore, the volume of turpentine remaining in the cylinder at 80 °C is 2.01 L .

Answer 14

(b)

To determine

The volume of turpentine remaining in the cylinder at 80 °C .

Answer 15

(b)

Expert Solution

Answer to Problem 19.21P

The volume of turpentine remaining in the cylinder at 80 °C is 2.01 L .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Given info: The initial volume of aluminum cylinder and turpentine is 2.000 L , the initial temperature is 20 °C , and the final temperature in the cylinder is 80 °C .

Write the expression for the new volume of turpentine in the cylinder,

Vnew=V(1+βtΔT)

The coefficient of volume expansion of turpentine is 9×10−4 (°C)−1 and from the part (a) the temperature difference is 60 °C .

Substitute 2.000 L for V , 9×10−4 (°C)−1 for βt and 60 °C for ΔT in the above equation.

Vnew=(2.000 L)[1+(9×10−4 (°C)−1)(60 °C)]=2.01 L

Conclusion:

Therefore, the volume of turpentine remaining in the cylinder at 80 °C is 2.01 L .

Answer 20

(c)

To determine

The distance for the cylinder’s rim does the turpentine surface recede.

(c)

Expert Solution

Answer to Problem 19.21P

The distance for the cylinder’s rim does the turpentine surface recede is 0.998 cm .

Explanation of Solution

Given info: The initial volume of aluminum cylinder and turpentine is 2.000 L , the final temperature is 20 °C , and the initial temperature in the cylinder is 80 °C .

Write the expression for the volume of the turpentine in the cylinder after it cools back to 20 °C ,

V1=Vnew(1+βtΔT′) (3)

Write the expression for temperature difference,

ΔT′=T2−T1

Here,

T2 is the final temperature.

T1 is the initial temperature.

Substitute 20 °C for T2 and 80 °C for T1 in the above equation.

ΔT′=20 °C−80 °C=−60 °C

Thus, the temperature difference is −60 °C and from part (a) the volume of turpentine remaining in the cylinder at 80 °C is 2.1 L .

Substitute −60 °C for ΔT′ , 2.1 L for Vnew and 9×10−4 °C−1 for βt in the equation (3).

V1=(2.1 L)(103 cm31 L)[1+(9×10−4 °C−1)(−60 °C)]=1900.17 cm3

Write the expression for the percentage of cylinder that is empty at 20 °C ,

%V=V−V1V

Substitute 1900.17 cm3 for V1 and 2.000 L for V in the above equation.

%V=(2.0 L)(103 cm31 L)−1900.17 cm3(2.0 L)(103 cm31 L)=4.99%

The distance for the cylinder’s rim does the turpentine surface recede is,

4.99%(20.0 cm)=0.998 cm

Conclusion:

Therefore, the distance for the cylinder’s rim does the turpentine surface recede is 0.998 cm .

Answer 21

(c)

To determine

The distance for the cylinder’s rim does the turpentine surface recede.

Answer 22

(c)

Expert Solution

Answer to Problem 19.21P

The distance for the cylinder’s rim does the turpentine surface recede is 0.998 cm .

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Given info: The initial volume of aluminum cylinder and turpentine is 2.000 L , the final temperature is 20 °C , and the initial temperature in the cylinder is 80 °C .