The amount of turpentine overflows.

Question

Want to see more full solutions like this?

Answer 1

Question

Chapter 19, Problem 21P

(a)

To determine

The amount of turpentine overflows.

(a)

Expert Solution

Answer to Problem 21P

The amount of turpentine overflows is 99.4 cm3_.

Explanation of Solution

Write the equation of volume that overflows.

(ΔV)=(β−3α)Vi(T2−T1) (I)

Here, (ΔV) is the amount of overflow liquid, β is the volume expansion coefficient, α is the linear expansion coefficient, Vi is the initial volume, T1 is the initial temperature and T2 is the final temperature.

Conclusion:

Substitute, 9.00×10−4/°C for β, 24.0×10−6/°C for α, 2.000 L for Vi, 20.0°C for T1 and 80.0°C for T2 in equation (I) to find (ΔV).

(ΔV)=([9.00×10−4/°C]−3[24.0×10−6/°C])(2.000 L)(80.0°C−20.0°C)=([9.00×10−4/°C]−3[24.0×10−6/°C])(2.000 L)(60.0°C)=99.4 L=99.4cm3

Thus, the amount of turpentine overflows is 99.4 cm3_.

(b)

To determine

The remaining volume of the turpentine in the cylinder.

(b)

Expert Solution

Answer to Problem 21P

The remaining volume of the turpentine in the cylinder is 2.01 L_.

Explanation of Solution

Write the equation of remaining volume of the liquid in the cylinder.

Vr=Vi+3αVi(T2−T1) (II)

Here, Vr is the remaining volume of liquid.

Conclusion:

Substitute, 24.0×10−6/°C for α, 2.000 L for Vi, 20.0°C for T1 and 80.0°C for T2 in equation (I) to find Vr.

Vr=(2.000 L)+3(24.0×10−6/°C)(2.000 L)(80.0°C−20.0°C)=(2.000 L)+3(24.0×10−6/°C)(2.000 L)(60.0°C)=2.01 L

Thus, the remaining volume of the turpentine in the cylinder is 2.01 L_.

(c)

To determine

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

(c)

Expert Solution

Answer to Problem 21P

The distance at which the cylinder’s rim does the turpentine’s surface recede is 0.998 cm_.

Explanation of Solution

Write the equation for the distance.

d=ph (III)

Here, d is the distance, p is the percentage of the empty cylinder and h is the depth of the cylinder.

Write the expression for the percentage.

p=(Vi−V'Vi)×100% (IV)

Here, V' is the volume of the liquid in the cylinder after it cools back to 20°C.

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

V'=V(1+β(T2−T1)) (V)

Rewrite the expression for the distance from equation (III) by using (IV) and (V).

d=[(Vi−[V(1+β(T2−T1))]Vi)×100%]h (VI)

Conclusion:

Substitute, 9.00×10−4/°C for β, 2.01 L for V, 2.000 L for Vi, 80.0°C for T1, 20.0 cm for h and 20.0°C for T2 in equation (I) to find (ΔV).

d=[((2.000 L)−[(2.01 L)(1+(9.00×10−4/°C)(20.0°C−80.0°C))](2.000 L))×100%](20.0 cm)=(4.99%)(20.0 cm)=0.998 cm

Therefore, the distance at which the cylinder’s rim does the turpentine’s surface recede is 0.998 cm_.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

ROTATIONAL DYNAMICS Question 01 A solid circular cylinder and a solid spherical ball of the same mass and radius are rolling together down the same inclined. Calculate the ratio of their kinetic energy. Assume pure rolling motion Question 02 A sphere and cylinder of the same mass and radius start from ret at the same point and more down the same plane inclined at 30° to the horizontal Which body gets the bottom first and what is its acceleration b) What angle of inclination of the plane is needed to give the slower body the same acceleration Question 03 i) Define the angular velocity of a rotating body and give its SI unit A car wheel has its angular velocity changing from 2rads to 30 rads seconds. If the radius of the wheel is 400mm. calculate ii) The angular acceleration iii) The tangential linear acceleration of a point on the rim of the wheel Question 04 in 20

Question B3 Consider the following FLRW spacetime: t2 ds² = -dt² + (dx² + dy²+ dz²), t2 where t is a constant. a) State whether this universe is spatially open, closed or flat. [2 marks] b) Determine the Hubble factor H(t), and represent it in a (roughly drawn) plot as a function of time t, starting at t = 0. [3 marks] c) Taking galaxy A to be located at (x, y, z) = (0,0,0), determine the proper distance to galaxy B located at (x, y, z) = (L, 0, 0). Determine the recessional velocity of galaxy B with respect to galaxy A. d) The Friedmann equations are 2 k 8πG а 4πG + a² (p+3p). 3 a 3 [5 marks] Use these equations to determine the energy density p(t) and the pressure p(t) for the FLRW spacetime specified at the top of the page. [5 marks] e) Given the result of question B3.d, state whether the FLRW universe in question is (i) radiation-dominated, (ii) matter-dominated, (iii) cosmological-constant-dominated, or (iv) none of the previous. Justify your answer. f) [5 marks] A conformally…

SECTION B Answer ONLY TWO questions in Section B [Expect to use one single-sided A4 page for each Section-B sub question.] Question B1 Consider the line element where w is a constant. ds²=-dt²+e2wt dx², a) Determine the components of the metric and of the inverse metric. [2 marks] b) Determine the Christoffel symbols. [See the Appendix of this document.] [10 marks] c) Write down the geodesic equations. [5 marks] d) Show that e2wt it is a constant of geodesic motion. [4 marks] e) Solve the geodesic equations for null geodesics. [4 marks]

Answer 2

Question

Chapter 19, Problem 21P

(a)

To determine

The amount of turpentine overflows.

(a)

Expert Solution

Answer to Problem 21P

The amount of turpentine overflows is 99.4 cm3_.

Explanation of Solution

Write the equation of volume that overflows.

(ΔV)=(β−3α)Vi(T2−T1) (I)

Here, (ΔV) is the amount of overflow liquid, β is the volume expansion coefficient, α is the linear expansion coefficient, Vi is the initial volume, T1 is the initial temperature and T2 is the final temperature.

Conclusion:

Substitute, 9.00×10−4/°C for β, 24.0×10−6/°C for α, 2.000 L for Vi, 20.0°C for T1 and 80.0°C for T2 in equation (I) to find (ΔV).

(ΔV)=([9.00×10−4/°C]−3[24.0×10−6/°C])(2.000 L)(80.0°C−20.0°C)=([9.00×10−4/°C]−3[24.0×10−6/°C])(2.000 L)(60.0°C)=99.4 L=99.4cm3

Thus, the amount of turpentine overflows is 99.4 cm3_.

(b)

To determine

The remaining volume of the turpentine in the cylinder.

(b)

Expert Solution

Answer to Problem 21P

The remaining volume of the turpentine in the cylinder is 2.01 L_.

Explanation of Solution

Write the equation of remaining volume of the liquid in the cylinder.

Vr=Vi+3αVi(T2−T1) (II)

Here, Vr is the remaining volume of liquid.

Conclusion:

Substitute, 24.0×10−6/°C for α, 2.000 L for Vi, 20.0°C for T1 and 80.0°C for T2 in equation (I) to find Vr.

Vr=(2.000 L)+3(24.0×10−6/°C)(2.000 L)(80.0°C−20.0°C)=(2.000 L)+3(24.0×10−6/°C)(2.000 L)(60.0°C)=2.01 L

Thus, the remaining volume of the turpentine in the cylinder is 2.01 L_.

(c)

To determine

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

(c)

Expert Solution

Answer to Problem 21P

The distance at which the cylinder’s rim does the turpentine’s surface recede is 0.998 cm_.

Explanation of Solution

Write the equation for the distance.

d=ph (III)

Here, d is the distance, p is the percentage of the empty cylinder and h is the depth of the cylinder.

Write the expression for the percentage.

p=(Vi−V'Vi)×100% (IV)

Here, V' is the volume of the liquid in the cylinder after it cools back to 20°C.

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

V'=V(1+β(T2−T1)) (V)

Rewrite the expression for the distance from equation (III) by using (IV) and (V).

d=[(Vi−[V(1+β(T2−T1))]Vi)×100%]h (VI)

Conclusion:

Substitute, 9.00×10−4/°C for β, 2.01 L for V, 2.000 L for Vi, 80.0°C for T1, 20.0 cm for h and 20.0°C for T2 in equation (I) to find (ΔV).

d=[((2.000 L)−[(2.01 L)(1+(9.00×10−4/°C)(20.0°C−80.0°C))](2.000 L))×100%](20.0 cm)=(4.99%)(20.0 cm)=0.998 cm

Therefore, the distance at which the cylinder’s rim does the turpentine’s surface recede is 0.998 cm_.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Question

Chapter 19, Problem 21P

(a)

To determine

The amount of turpentine overflows.

(a)

Expert Solution

Answer to Problem 21P

The amount of turpentine overflows is 99.4 cm3_.

Explanation of Solution

Write the equation of volume that overflows.

(ΔV)=(β−3α)Vi(T2−T1) (I)

Here, (ΔV) is the amount of overflow liquid, β is the volume expansion coefficient, α is the linear expansion coefficient, Vi is the initial volume, T1 is the initial temperature and T2 is the final temperature.

Conclusion:

Substitute, 9.00×10−4/°C for β, 24.0×10−6/°C for α, 2.000 L for Vi, 20.0°C for T1 and 80.0°C for T2 in equation (I) to find (ΔV).

(ΔV)=([9.00×10−4/°C]−3[24.0×10−6/°C])(2.000 L)(80.0°C−20.0°C)=([9.00×10−4/°C]−3[24.0×10−6/°C])(2.000 L)(60.0°C)=99.4 L=99.4cm3

Thus, the amount of turpentine overflows is 99.4 cm3_.

(b)

To determine

The remaining volume of the turpentine in the cylinder.

(b)

Expert Solution

Answer to Problem 21P

The remaining volume of the turpentine in the cylinder is 2.01 L_.

Explanation of Solution

Write the equation of remaining volume of the liquid in the cylinder.

Vr=Vi+3αVi(T2−T1) (II)

Here, Vr is the remaining volume of liquid.

Conclusion:

Substitute, 24.0×10−6/°C for α, 2.000 L for Vi, 20.0°C for T1 and 80.0°C for T2 in equation (I) to find Vr.

Vr=(2.000 L)+3(24.0×10−6/°C)(2.000 L)(80.0°C−20.0°C)=(2.000 L)+3(24.0×10−6/°C)(2.000 L)(60.0°C)=2.01 L

Thus, the remaining volume of the turpentine in the cylinder is 2.01 L_.

(c)

To determine

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

(c)

Expert Solution

Answer to Problem 21P

The distance at which the cylinder’s rim does the turpentine’s surface recede is 0.998 cm_.

Explanation of Solution

Write the equation for the distance.

d=ph (III)

Here, d is the distance, p is the percentage of the empty cylinder and h is the depth of the cylinder.

Write the expression for the percentage.

p=(Vi−V'Vi)×100% (IV)

Here, V' is the volume of the liquid in the cylinder after it cools back to 20°C.

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

V'=V(1+β(T2−T1)) (V)

Rewrite the expression for the distance from equation (III) by using (IV) and (V).

d=[(Vi−[V(1+β(T2−T1))]Vi)×100%]h (VI)

Conclusion:

Substitute, 9.00×10−4/°C for β, 2.01 L for V, 2.000 L for Vi, 80.0°C for T1, 20.0 cm for h and 20.0°C for T2 in equation (I) to find (ΔV).

d=[((2.000 L)−[(2.01 L)(1+(9.00×10−4/°C)(20.0°C−80.0°C))](2.000 L))×100%](20.0 cm)=(4.99%)(20.0 cm)=0.998 cm

Therefore, the distance at which the cylinder’s rim does the turpentine’s surface recede is 0.998 cm_.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Question

Chapter 19, Problem 21P

(a)

To determine

The amount of turpentine overflows.

(a)

Expert Solution

Answer to Problem 21P

The amount of turpentine overflows is 99.4 cm3_.

Explanation of Solution

Write the equation of volume that overflows.

(ΔV)=(β−3α)Vi(T2−T1) (I)

Here, (ΔV) is the amount of overflow liquid, β is the volume expansion coefficient, α is the linear expansion coefficient, Vi is the initial volume, T1 is the initial temperature and T2 is the final temperature.

Conclusion:

Substitute, 9.00×10−4/°C for β, 24.0×10−6/°C for α, 2.000 L for Vi, 20.0°C for T1 and 80.0°C for T2 in equation (I) to find (ΔV).

(ΔV)=([9.00×10−4/°C]−3[24.0×10−6/°C])(2.000 L)(80.0°C−20.0°C)=([9.00×10−4/°C]−3[24.0×10−6/°C])(2.000 L)(60.0°C)=99.4 L=99.4cm3

Thus, the amount of turpentine overflows is 99.4 cm3_.

(b)

To determine

The remaining volume of the turpentine in the cylinder.

(b)

Expert Solution

Answer to Problem 21P

The remaining volume of the turpentine in the cylinder is 2.01 L_.

Explanation of Solution

Write the equation of remaining volume of the liquid in the cylinder.

Vr=Vi+3αVi(T2−T1) (II)

Here, Vr is the remaining volume of liquid.

Conclusion:

Substitute, 24.0×10−6/°C for α, 2.000 L for Vi, 20.0°C for T1 and 80.0°C for T2 in equation (I) to find Vr.

Vr=(2.000 L)+3(24.0×10−6/°C)(2.000 L)(80.0°C−20.0°C)=(2.000 L)+3(24.0×10−6/°C)(2.000 L)(60.0°C)=2.01 L

Thus, the remaining volume of the turpentine in the cylinder is 2.01 L_.

(c)

To determine

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

(c)

Expert Solution

Answer to Problem 21P

The distance at which the cylinder’s rim does the turpentine’s surface recede is 0.998 cm_.

Explanation of Solution

Write the equation for the distance.

d=ph (III)

Here, d is the distance, p is the percentage of the empty cylinder and h is the depth of the cylinder.

Write the expression for the percentage.

p=(Vi−V'Vi)×100% (IV)

Here, V' is the volume of the liquid in the cylinder after it cools back to 20°C.

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

V'=V(1+β(T2−T1)) (V)

Rewrite the expression for the distance from equation (III) by using (IV) and (V).

d=[(Vi−[V(1+β(T2−T1))]Vi)×100%]h (VI)

Conclusion:

Substitute, 9.00×10−4/°C for β, 2.01 L for V, 2.000 L for Vi, 80.0°C for T1, 20.0 cm for h and 20.0°C for T2 in equation (I) to find (ΔV).

d=[((2.000 L)−[(2.01 L)(1+(9.00×10−4/°C)(20.0°C−80.0°C))](2.000 L))×100%](20.0 cm)=(4.99%)(20.0 cm)=0.998 cm

Therefore, the distance at which the cylinder’s rim does the turpentine’s surface recede is 0.998 cm_.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 19, Problem 21P

(a)

To determine

The amount of turpentine overflows.

(a)

Expert Solution

Answer to Problem 21P

The amount of turpentine overflows is 99.4 cm3_.

Explanation of Solution

Write the equation of volume that overflows.

(ΔV)=(β−3α)Vi(T2−T1) (I)

Here, (ΔV) is the amount of overflow liquid, β is the volume expansion coefficient, α is the linear expansion coefficient, Vi is the initial volume, T1 is the initial temperature and T2 is the final temperature.

Conclusion:

Substitute, 9.00×10−4/°C for β, 24.0×10−6/°C for α, 2.000 L for Vi, 20.0°C for T1 and 80.0°C for T2 in equation (I) to find (ΔV).

(ΔV)=([9.00×10−4/°C]−3[24.0×10−6/°C])(2.000 L)(80.0°C−20.0°C)=([9.00×10−4/°C]−3[24.0×10−6/°C])(2.000 L)(60.0°C)=99.4 L=99.4cm3

Thus, the amount of turpentine overflows is 99.4 cm3_.

(b)

To determine

The remaining volume of the turpentine in the cylinder.

(b)

Expert Solution

Answer to Problem 21P

The remaining volume of the turpentine in the cylinder is 2.01 L_.

Explanation of Solution

Write the equation of remaining volume of the liquid in the cylinder.

Vr=Vi+3αVi(T2−T1) (II)

Here, Vr is the remaining volume of liquid.

Conclusion:

Substitute, 24.0×10−6/°C for α, 2.000 L for Vi, 20.0°C for T1 and 80.0°C for T2 in equation (I) to find Vr.

Vr=(2.000 L)+3(24.0×10−6/°C)(2.000 L)(80.0°C−20.0°C)=(2.000 L)+3(24.0×10−6/°C)(2.000 L)(60.0°C)=2.01 L

Thus, the remaining volume of the turpentine in the cylinder is 2.01 L_.

(c)

To determine

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

(c)

Expert Solution

Answer to Problem 21P

The distance at which the cylinder’s rim does the turpentine’s surface recede is 0.998 cm_.

Explanation of Solution

Write the equation for the distance.

d=ph (III)

Here, d is the distance, p is the percentage of the empty cylinder and h is the depth of the cylinder.

Write the expression for the percentage.

p=(Vi−V'Vi)×100% (IV)

Here, V' is the volume of the liquid in the cylinder after it cools back to 20°C.

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

V'=V(1+β(T2−T1)) (V)

Rewrite the expression for the distance from equation (III) by using (IV) and (V).

d=[(Vi−[V(1+β(T2−T1))]Vi)×100%]h (VI)

Conclusion:

Substitute, 9.00×10−4/°C for β, 2.01 L for V, 2.000 L for Vi, 80.0°C for T1, 20.0 cm for h and 20.0°C for T2 in equation (I) to find (ΔV).

d=[((2.000 L)−[(2.01 L)(1+(9.00×10−4/°C)(20.0°C−80.0°C))](2.000 L))×100%](20.0 cm)=(4.99%)(20.0 cm)=0.998 cm

Therefore, the distance at which the cylinder’s rim does the turpentine’s surface recede is 0.998 cm_.

Answer 6

Chapter 19, Problem 21P

(a)

To determine

The amount of turpentine overflows.

(a)

Expert Solution

Answer to Problem 21P

The amount of turpentine overflows is 99.4 cm3_.

Explanation of Solution

Write the equation of volume that overflows.

(ΔV)=(β−3α)Vi(T2−T1) (I)

Here, (ΔV) is the amount of overflow liquid, β is the volume expansion coefficient, α is the linear expansion coefficient, Vi is the initial volume, T1 is the initial temperature and T2 is the final temperature.

Conclusion:

Substitute, 9.00×10−4/°C for β, 24.0×10−6/°C for α, 2.000 L for Vi, 20.0°C for T1 and 80.0°C for T2 in equation (I) to find (ΔV).

(ΔV)=([9.00×10−4/°C]−3[24.0×10−6/°C])(2.000 L)(80.0°C−20.0°C)=([9.00×10−4/°C]−3[24.0×10−6/°C])(2.000 L)(60.0°C)=99.4 L=99.4cm3

Thus, the amount of turpentine overflows is 99.4 cm3_.

Answer 7

Chapter 19, Problem 21P

(a)

To determine

The amount of turpentine overflows.

Answer 8

(a)

Expert Solution

Answer to Problem 21P

The amount of turpentine overflows is 99.4 cm3_.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Write the equation of volume that overflows.

(ΔV)=(β−3α)Vi(T2−T1) (I)

Here, (ΔV) is the amount of overflow liquid, β is the volume expansion coefficient, α is the linear expansion coefficient, Vi is the initial volume, T1 is the initial temperature and T2 is the final temperature.

Conclusion:

Substitute, 9.00×10−4/°C for β, 24.0×10−6/°C for α, 2.000 L for Vi, 20.0°C for T1 and 80.0°C for T2 in equation (I) to find (ΔV).

(ΔV)=([9.00×10−4/°C]−3[24.0×10−6/°C])(2.000 L)(80.0°C−20.0°C)=([9.00×10−4/°C]−3[24.0×10−6/°C])(2.000 L)(60.0°C)=99.4 L=99.4cm3

Thus, the amount of turpentine overflows is 99.4 cm3_.

Answer 13

(b)

To determine

The remaining volume of the turpentine in the cylinder.

(b)

Expert Solution

Answer to Problem 21P

The remaining volume of the turpentine in the cylinder is 2.01 L_.

Explanation of Solution

Write the equation of remaining volume of the liquid in the cylinder.

Vr=Vi+3αVi(T2−T1) (II)

Here, Vr is the remaining volume of liquid.

Conclusion:

Substitute, 24.0×10−6/°C for α, 2.000 L for Vi, 20.0°C for T1 and 80.0°C for T2 in equation (I) to find Vr.

Vr=(2.000 L)+3(24.0×10−6/°C)(2.000 L)(80.0°C−20.0°C)=(2.000 L)+3(24.0×10−6/°C)(2.000 L)(60.0°C)=2.01 L

Thus, the remaining volume of the turpentine in the cylinder is 2.01 L_.

Answer 14

(b)

To determine

The remaining volume of the turpentine in the cylinder.

Answer 15

(b)

Expert Solution

Answer to Problem 21P

The remaining volume of the turpentine in the cylinder is 2.01 L_.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Write the equation of remaining volume of the liquid in the cylinder.

Vr=Vi+3αVi(T2−T1) (II)

Here, Vr is the remaining volume of liquid.

Conclusion:

Substitute, 24.0×10−6/°C for α, 2.000 L for Vi, 20.0°C for T1 and 80.0°C for T2 in equation (I) to find Vr.

Vr=(2.000 L)+3(24.0×10−6/°C)(2.000 L)(80.0°C−20.0°C)=(2.000 L)+3(24.0×10−6/°C)(2.000 L)(60.0°C)=2.01 L

Thus, the remaining volume of the turpentine in the cylinder is 2.01 L_.

Answer 20

(c)

To determine

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

(c)

Expert Solution

Answer to Problem 21P

The distance at which the cylinder’s rim does the turpentine’s surface recede is 0.998 cm_.

Explanation of Solution

Write the equation for the distance.

d=ph (III)

Here, d is the distance, p is the percentage of the empty cylinder and h is the depth of the cylinder.

Write the expression for the percentage.

p=(Vi−V'Vi)×100% (IV)

Here, V' is the volume of the liquid in the cylinder after it cools back to 20°C.

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

V'=V(1+β(T2−T1)) (V)

Rewrite the expression for the distance from equation (III) by using (IV) and (V).

d=[(Vi−[V(1+β(T2−T1))]Vi)×100%]h (VI)

Conclusion:

Substitute, 9.00×10−4/°C for β, 2.01 L for V, 2.000 L for Vi, 80.0°C for T1, 20.0 cm for h and 20.0°C for T2 in equation (I) to find (ΔV).

d=[((2.000 L)−[(2.01 L)(1+(9.00×10−4/°C)(20.0°C−80.0°C))](2.000 L))×100%](20.0 cm)=(4.99%)(20.0 cm)=0.998 cm

Therefore, the distance at which the cylinder’s rim does the turpentine’s surface recede is 0.998 cm_.

Answer 21

(c)

To determine

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

Answer 22

(c)

Expert Solution

Answer to Problem 21P

The distance at which the cylinder’s rim does the turpentine’s 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

Write the equation for the distance.

d=ph (III)

Here, d is the distance, p is the percentage of the empty cylinder and h is the depth of the cylinder.

Write the expression for the percentage.

p=(Vi−V'Vi)×100% (IV)

Here, V' is the volume of the liquid in the cylinder after it cools back to 20°C.

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

V'=V(1+β(T2−T1)) (V)

Rewrite the expression for the distance from equation (III) by using (IV) and (V).

d=[(Vi−[V(1+β(T2−T1))]Vi)×100%]h (VI)

Conclusion:

Substitute, 9.00×10−4/°C for β, 2.01 L for V, 2.000 L for Vi, 80.0°C for T1, 20.0 cm for h and 20.0°C for T2 in equation (I) to find (ΔV).

d=[((2.000 L)−[(2.01 L)(1+(9.00×10−4/°C)(20.0°C−80.0°C))](2.000 L))×100%](20.0 cm)=(4.99%)(20.0 cm)=0.998 cm

Therefore, the distance at which the cylinder’s rim does the turpentine’s surface recede is 0.998 cm_.

Answer 27

Ch. 19.1 - Prob. 19.1QQ Ch. 19.3 - Consider the following pairs of materials. Which...Ch. 19.4 - If you are asked to make a very sensitive glass...Ch. 19.4 - Prob. 19.4QQ Ch. 19.5 - A common material for cushioning objects in...Ch. 19.5 - On a winter day, you turn on your furnace and the...Ch. 19 - Prob. 1OQ Ch. 19 - Prob. 2OQ Ch. 19 - Prob. 3OQ Ch. 19 - Prob. 4OQ

The amount of turpentine overflows.

Concept explainers

Videos

The amount of turpentine overflows.

Answer to Problem 21P

Explanation of Solution

The remaining volume of the turpentine in the cylinder.

Answer to Problem 21P

Explanation of Solution

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

Answer to Problem 21P

Explanation of Solution

Want to see more full solutions like this?

Chapter 19 Solutions