The fraction change in density for a change in temperature is given by Δ ρ ρ = − β Δ T .

Question

Want to see more full solutions like this?

Answer 1

Question

Chapter 16, Problem 62P

(a)

To determine

To show: The fraction change in density for a change in temperature is given by Δρρ=−βΔT .

(a)

Expert Solution

Answer to Problem 62P

The fraction change in density for a change in temperature is Δρρ=−βΔT .

Explanation of Solution

Given info: The density of the liquid is ρ , the change in the temperature is ΔT , the maximum density of fresh water at 4.0 °C is 1.0000 g/cm3 , the density of fresh water at 10.0 °C is 0.9997 g/cm3 and the density of water at 0 °C is 0.9999 g/cm3 .

The expression for the coefficient of the volume expansion is,

βΔT=ΔVV

Here,

β is the coefficient of the volume expansion.

The formula for the density is,

ρ=mV

Here,

m is the mass of the water.

V is the volume of the water.\

Differentiate the above equation with respect to volume.

dρ=−mV2dV

For very small change in volume and density the expression is,

Δρ=−mV2ΔV=−mVΔVV

Substitute ρ for mV and βΔT for ΔVV in above equation.

Δρ=−ρβΔTΔρρ=−βΔT

Conclusion:

Therefore, the fraction change in density for a change in temperature is Δρρ=−βΔT .

(b)

To determine

The significance of the negative sign.

(b)

Expert Solution

Answer to Problem 62P

the negative sign signifies that when the temperature increases, the density decreases.

Explanation of Solution

Given info: The density of the liquid is ρ , the change in the temperature is ΔT , the maximum density of fresh water at 4.0 °C is 1.0000 g/cm3 , the density of fresh water at 10.0 °C is 0.9997 g/cm3 and the density of water at 0 °C is 0.9999 g/cm3 .

The expression for the fraction of the density is,

Δρρ=−βΔT (1)

The negative sign in the above expression shows that as the temperature increases, the density of the water is decreases.

Conclusion:

Therefore, the negative sign signifies that when the temperature increases, the density decreases.

(c)

To determine

The value of β for the temperature interval 4.0 °C to 10.0 °C .

(c)

Expert Solution

Answer to Problem 62P

The value of β for the temperature interval 4.0 °C to 10.0 °C is 5×10−5 (°C)−1 .

Explanation of Solution

Given info: The density of the liquid is ρ , the change in the temperature is ΔT , the maximum density of fresh water at 4.0 °C is 1.0000 g/cm3 , the density of fresh water at 10.0 °C is 0.9997 g/cm3 and the density of water at 0 °C is 0.9999 g/cm3 .

Recall the equation (1) and rearrange for β .

Δρρ=−βΔTβ=−ΔρρΔT

Substitute (0.9997 g/cm3−1.0000 g/cm3) for Δρ , 1.0000 g/cm3 for ρ and (10.0 °C−4.0 °C) for ΔT in above equation to find β .

β=−(0.9997 g/cm3−1.0000 g/cm3)(1.0000 g/cm3)((10.0 °C−4.0 °C))=5×10−5 °C−1

Conclusion:

Therefore, the value of β for temperature interval 4.0 °C to 10.0 °C is 5×10−5 (°C)−1 .

(d)

To determine

The value of β for the temperature interval 0 °C to 4.0 °C .

(d)

Expert Solution

Answer to Problem 62P

The value of β for temperature interval 0 °C to 4.0 °C is −2.5×10−5 (°C)−1 .

Explanation of Solution

Given info: The density of the liquid is ρ , the change in the temperature is ΔT , the maximum density of fresh water at 4.0 °C is 1.0000 g/cm3 , the density of fresh water at 10.0 °C is 0.9997 g/cm3 and the density of water at 0 °C is 0.9999 g/cm3 .

The expression for β is,

Δρρ=−βΔTβ=−ΔρρΔT

Substitute (1.0000 g/cm3−0.9999 g/cm3) for Δρ , 1.0000 g/cm3 for ρ and (4.0 °C−0 °C) for ΔT in above equation to find β .

β=−(1.0000 g/cm3−0.9999 g/cm3)(1.0000 g/cm3)((4.0 °C−0.0 °C))=−2.5×10−5 °C−1

Conclusion:

Therefore, the value of β for the temperature interval 0 °C to 4.0 °C is −2.5×10−5 (°C)−1 .

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

No chatgpt pls will upvote

suggest a reason ultrasound cleaning is better than cleaning by hand?

Checkpoint 4 The figure shows four orientations of an electric di- pole in an external electric field. Rank the orienta- tions according to (a) the magnitude of the torque on the dipole and (b) the potential energy of the di- pole, greatest first. (1) (2) E (4)

Answer 2

Question

Chapter 16, Problem 62P

(a)

To determine

To show: The fraction change in density for a change in temperature is given by Δρρ=−βΔT .

(a)

Expert Solution

Answer to Problem 62P

The fraction change in density for a change in temperature is Δρρ=−βΔT .

Explanation of Solution

Given info: The density of the liquid is ρ , the change in the temperature is ΔT , the maximum density of fresh water at 4.0 °C is 1.0000 g/cm3 , the density of fresh water at 10.0 °C is 0.9997 g/cm3 and the density of water at 0 °C is 0.9999 g/cm3 .

The expression for the coefficient of the volume expansion is,

βΔT=ΔVV

Here,

β is the coefficient of the volume expansion.

The formula for the density is,

ρ=mV

Here,

m is the mass of the water.

V is the volume of the water.\

Differentiate the above equation with respect to volume.

dρ=−mV2dV

For very small change in volume and density the expression is,

Δρ=−mV2ΔV=−mVΔVV

Substitute ρ for mV and βΔT for ΔVV in above equation.

Δρ=−ρβΔTΔρρ=−βΔT

Conclusion:

Therefore, the fraction change in density for a change in temperature is Δρρ=−βΔT .

(b)

To determine

The significance of the negative sign.

(b)

Expert Solution

Answer to Problem 62P

the negative sign signifies that when the temperature increases, the density decreases.

Explanation of Solution

Given info: The density of the liquid is ρ , the change in the temperature is ΔT , the maximum density of fresh water at 4.0 °C is 1.0000 g/cm3 , the density of fresh water at 10.0 °C is 0.9997 g/cm3 and the density of water at 0 °C is 0.9999 g/cm3 .

The expression for the fraction of the density is,

Δρρ=−βΔT (1)

The negative sign in the above expression shows that as the temperature increases, the density of the water is decreases.

Conclusion:

Therefore, the negative sign signifies that when the temperature increases, the density decreases.

(c)

To determine

The value of β for the temperature interval 4.0 °C to 10.0 °C .

(c)

Expert Solution

Answer to Problem 62P

The value of β for the temperature interval 4.0 °C to 10.0 °C is 5×10−5 (°C)−1 .

Explanation of Solution

Given info: The density of the liquid is ρ , the change in the temperature is ΔT , the maximum density of fresh water at 4.0 °C is 1.0000 g/cm3 , the density of fresh water at 10.0 °C is 0.9997 g/cm3 and the density of water at 0 °C is 0.9999 g/cm3 .

Recall the equation (1) and rearrange for β .

Δρρ=−βΔTβ=−ΔρρΔT

Substitute (0.9997 g/cm3−1.0000 g/cm3) for Δρ , 1.0000 g/cm3 for ρ and (10.0 °C−4.0 °C) for ΔT in above equation to find β .

β=−(0.9997 g/cm3−1.0000 g/cm3)(1.0000 g/cm3)((10.0 °C−4.0 °C))=5×10−5 °C−1

Conclusion:

Therefore, the value of β for temperature interval 4.0 °C to 10.0 °C is 5×10−5 (°C)−1 .

(d)

To determine

The value of β for the temperature interval 0 °C to 4.0 °C .

(d)

Expert Solution

Answer to Problem 62P

The value of β for temperature interval 0 °C to 4.0 °C is −2.5×10−5 (°C)−1 .

Explanation of Solution

Given info: The density of the liquid is ρ , the change in the temperature is ΔT , the maximum density of fresh water at 4.0 °C is 1.0000 g/cm3 , the density of fresh water at 10.0 °C is 0.9997 g/cm3 and the density of water at 0 °C is 0.9999 g/cm3 .

The expression for β is,

Δρρ=−βΔTβ=−ΔρρΔT

Substitute (1.0000 g/cm3−0.9999 g/cm3) for Δρ , 1.0000 g/cm3 for ρ and (4.0 °C−0 °C) for ΔT in above equation to find β .

β=−(1.0000 g/cm3−0.9999 g/cm3)(1.0000 g/cm3)((4.0 °C−0.0 °C))=−2.5×10−5 °C−1

Conclusion:

Therefore, the value of β for the temperature interval 0 °C to 4.0 °C is −2.5×10−5 (°C)−1 .

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 16, Problem 62P

(a)

To determine

To show: The fraction change in density for a change in temperature is given by Δρρ=−βΔT .

(a)

Expert Solution

Answer to Problem 62P

The fraction change in density for a change in temperature is Δρρ=−βΔT .

Explanation of Solution

Given info: The density of the liquid is ρ , the change in the temperature is ΔT , the maximum density of fresh water at 4.0 °C is 1.0000 g/cm3 , the density of fresh water at 10.0 °C is 0.9997 g/cm3 and the density of water at 0 °C is 0.9999 g/cm3 .

The expression for the coefficient of the volume expansion is,

βΔT=ΔVV

Here,

β is the coefficient of the volume expansion.

The formula for the density is,

ρ=mV

Here,

m is the mass of the water.

V is the volume of the water.\

Differentiate the above equation with respect to volume.

dρ=−mV2dV

For very small change in volume and density the expression is,

Δρ=−mV2ΔV=−mVΔVV

Substitute ρ for mV and βΔT for ΔVV in above equation.

Δρ=−ρβΔTΔρρ=−βΔT

Conclusion:

Therefore, the fraction change in density for a change in temperature is Δρρ=−βΔT .

(b)

To determine

The significance of the negative sign.

(b)

Expert Solution

Answer to Problem 62P

the negative sign signifies that when the temperature increases, the density decreases.

Explanation of Solution

Given info: The density of the liquid is ρ , the change in the temperature is ΔT , the maximum density of fresh water at 4.0 °C is 1.0000 g/cm3 , the density of fresh water at 10.0 °C is 0.9997 g/cm3 and the density of water at 0 °C is 0.9999 g/cm3 .

The expression for the fraction of the density is,

Δρρ=−βΔT (1)

The negative sign in the above expression shows that as the temperature increases, the density of the water is decreases.

Conclusion:

Therefore, the negative sign signifies that when the temperature increases, the density decreases.

(c)

To determine

The value of β for the temperature interval 4.0 °C to 10.0 °C .

(c)

Expert Solution

Answer to Problem 62P

The value of β for the temperature interval 4.0 °C to 10.0 °C is 5×10−5 (°C)−1 .

Explanation of Solution

Given info: The density of the liquid is ρ , the change in the temperature is ΔT , the maximum density of fresh water at 4.0 °C is 1.0000 g/cm3 , the density of fresh water at 10.0 °C is 0.9997 g/cm3 and the density of water at 0 °C is 0.9999 g/cm3 .

Recall the equation (1) and rearrange for β .

Δρρ=−βΔTβ=−ΔρρΔT

Substitute (0.9997 g/cm3−1.0000 g/cm3) for Δρ , 1.0000 g/cm3 for ρ and (10.0 °C−4.0 °C) for ΔT in above equation to find β .

β=−(0.9997 g/cm3−1.0000 g/cm3)(1.0000 g/cm3)((10.0 °C−4.0 °C))=5×10−5 °C−1

Conclusion:

Therefore, the value of β for temperature interval 4.0 °C to 10.0 °C is 5×10−5 (°C)−1 .

(d)

To determine

The value of β for the temperature interval 0 °C to 4.0 °C .

(d)

Expert Solution

Answer to Problem 62P

The value of β for temperature interval 0 °C to 4.0 °C is −2.5×10−5 (°C)−1 .

Explanation of Solution

Given info: The density of the liquid is ρ , the change in the temperature is ΔT , the maximum density of fresh water at 4.0 °C is 1.0000 g/cm3 , the density of fresh water at 10.0 °C is 0.9997 g/cm3 and the density of water at 0 °C is 0.9999 g/cm3 .

The expression for β is,

Δρρ=−βΔTβ=−ΔρρΔT

Substitute (1.0000 g/cm3−0.9999 g/cm3) for Δρ , 1.0000 g/cm3 for ρ and (4.0 °C−0 °C) for ΔT in above equation to find β .

β=−(1.0000 g/cm3−0.9999 g/cm3)(1.0000 g/cm3)((4.0 °C−0.0 °C))=−2.5×10−5 °C−1

Conclusion:

Therefore, the value of β for the temperature interval 0 °C to 4.0 °C is −2.5×10−5 (°C)−1 .

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 16, Problem 62P

(a)

To determine

To show: The fraction change in density for a change in temperature is given by Δρρ=−βΔT .

(a)

Expert Solution

Answer to Problem 62P

The fraction change in density for a change in temperature is Δρρ=−βΔT .

Explanation of Solution

Given info: The density of the liquid is ρ , the change in the temperature is ΔT , the maximum density of fresh water at 4.0 °C is 1.0000 g/cm3 , the density of fresh water at 10.0 °C is 0.9997 g/cm3 and the density of water at 0 °C is 0.9999 g/cm3 .

The expression for the coefficient of the volume expansion is,

βΔT=ΔVV

Here,

β is the coefficient of the volume expansion.

The formula for the density is,

ρ=mV

Here,

m is the mass of the water.

V is the volume of the water.\

Differentiate the above equation with respect to volume.

dρ=−mV2dV

For very small change in volume and density the expression is,

Δρ=−mV2ΔV=−mVΔVV

Substitute ρ for mV and βΔT for ΔVV in above equation.

Δρ=−ρβΔTΔρρ=−βΔT

Conclusion:

Therefore, the fraction change in density for a change in temperature is Δρρ=−βΔT .

(b)

To determine

The significance of the negative sign.

(b)

Expert Solution

Answer to Problem 62P

the negative sign signifies that when the temperature increases, the density decreases.

Explanation of Solution

Given info: The density of the liquid is ρ , the change in the temperature is ΔT , the maximum density of fresh water at 4.0 °C is 1.0000 g/cm3 , the density of fresh water at 10.0 °C is 0.9997 g/cm3 and the density of water at 0 °C is 0.9999 g/cm3 .

The expression for the fraction of the density is,

Δρρ=−βΔT (1)

The negative sign in the above expression shows that as the temperature increases, the density of the water is decreases.

Conclusion:

Therefore, the negative sign signifies that when the temperature increases, the density decreases.

(c)

To determine

The value of β for the temperature interval 4.0 °C to 10.0 °C .

(c)

Expert Solution

Answer to Problem 62P

The value of β for the temperature interval 4.0 °C to 10.0 °C is 5×10−5 (°C)−1 .

Explanation of Solution

Given info: The density of the liquid is ρ , the change in the temperature is ΔT , the maximum density of fresh water at 4.0 °C is 1.0000 g/cm3 , the density of fresh water at 10.0 °C is 0.9997 g/cm3 and the density of water at 0 °C is 0.9999 g/cm3 .

Recall the equation (1) and rearrange for β .

Δρρ=−βΔTβ=−ΔρρΔT

Substitute (0.9997 g/cm3−1.0000 g/cm3) for Δρ , 1.0000 g/cm3 for ρ and (10.0 °C−4.0 °C) for ΔT in above equation to find β .

β=−(0.9997 g/cm3−1.0000 g/cm3)(1.0000 g/cm3)((10.0 °C−4.0 °C))=5×10−5 °C−1

Conclusion:

Therefore, the value of β for temperature interval 4.0 °C to 10.0 °C is 5×10−5 (°C)−1 .

(d)

To determine

The value of β for the temperature interval 0 °C to 4.0 °C .

(d)

Expert Solution

Answer to Problem 62P

The value of β for temperature interval 0 °C to 4.0 °C is −2.5×10−5 (°C)−1 .

Explanation of Solution

Given info: The density of the liquid is ρ , the change in the temperature is ΔT , the maximum density of fresh water at 4.0 °C is 1.0000 g/cm3 , the density of fresh water at 10.0 °C is 0.9997 g/cm3 and the density of water at 0 °C is 0.9999 g/cm3 .

The expression for β is,

Δρρ=−βΔTβ=−ΔρρΔT

Substitute (1.0000 g/cm3−0.9999 g/cm3) for Δρ , 1.0000 g/cm3 for ρ and (4.0 °C−0 °C) for ΔT in above equation to find β .

β=−(1.0000 g/cm3−0.9999 g/cm3)(1.0000 g/cm3)((4.0 °C−0.0 °C))=−2.5×10−5 °C−1

Conclusion:

Therefore, the value of β for the temperature interval 0 °C to 4.0 °C is −2.5×10−5 (°C)−1 .

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 16, Problem 62P

(a)

To determine

To show: The fraction change in density for a change in temperature is given by Δρρ=−βΔT .

(a)

Expert Solution

Answer to Problem 62P

The fraction change in density for a change in temperature is Δρρ=−βΔT .

Explanation of Solution

Given info: The density of the liquid is ρ , the change in the temperature is ΔT , the maximum density of fresh water at 4.0 °C is 1.0000 g/cm3 , the density of fresh water at 10.0 °C is 0.9997 g/cm3 and the density of water at 0 °C is 0.9999 g/cm3 .

The expression for the coefficient of the volume expansion is,

βΔT=ΔVV

Here,

β is the coefficient of the volume expansion.

The formula for the density is,

ρ=mV

Here,

m is the mass of the water.

V is the volume of the water.\

Differentiate the above equation with respect to volume.

dρ=−mV2dV

For very small change in volume and density the expression is,

Δρ=−mV2ΔV=−mVΔVV

Substitute ρ for mV and βΔT for ΔVV in above equation.

Δρ=−ρβΔTΔρρ=−βΔT

Conclusion:

Therefore, the fraction change in density for a change in temperature is Δρρ=−βΔT .

(b)

To determine

The significance of the negative sign.

(b)

Expert Solution

Answer to Problem 62P

the negative sign signifies that when the temperature increases, the density decreases.

Explanation of Solution

Given info: The density of the liquid is ρ , the change in the temperature is ΔT , the maximum density of fresh water at 4.0 °C is 1.0000 g/cm3 , the density of fresh water at 10.0 °C is 0.9997 g/cm3 and the density of water at 0 °C is 0.9999 g/cm3 .

The expression for the fraction of the density is,

Δρρ=−βΔT (1)

The negative sign in the above expression shows that as the temperature increases, the density of the water is decreases.

Conclusion:

Therefore, the negative sign signifies that when the temperature increases, the density decreases.

(c)

To determine

The value of β for the temperature interval 4.0 °C to 10.0 °C .

(c)

Expert Solution

Answer to Problem 62P

The value of β for the temperature interval 4.0 °C to 10.0 °C is 5×10−5 (°C)−1 .

Explanation of Solution

Given info: The density of the liquid is ρ , the change in the temperature is ΔT , the maximum density of fresh water at 4.0 °C is 1.0000 g/cm3 , the density of fresh water at 10.0 °C is 0.9997 g/cm3 and the density of water at 0 °C is 0.9999 g/cm3 .

Recall the equation (1) and rearrange for β .

Δρρ=−βΔTβ=−ΔρρΔT

Substitute (0.9997 g/cm3−1.0000 g/cm3) for Δρ , 1.0000 g/cm3 for ρ and (10.0 °C−4.0 °C) for ΔT in above equation to find β .

β=−(0.9997 g/cm3−1.0000 g/cm3)(1.0000 g/cm3)((10.0 °C−4.0 °C))=5×10−5 °C−1

Conclusion:

Therefore, the value of β for temperature interval 4.0 °C to 10.0 °C is 5×10−5 (°C)−1 .

(d)

To determine

The value of β for the temperature interval 0 °C to 4.0 °C .

(d)

Expert Solution

Answer to Problem 62P

The value of β for temperature interval 0 °C to 4.0 °C is −2.5×10−5 (°C)−1 .

Explanation of Solution

Given info: The density of the liquid is ρ , the change in the temperature is ΔT , the maximum density of fresh water at 4.0 °C is 1.0000 g/cm3 , the density of fresh water at 10.0 °C is 0.9997 g/cm3 and the density of water at 0 °C is 0.9999 g/cm3 .

The expression for β is,

Δρρ=−βΔTβ=−ΔρρΔT

Substitute (1.0000 g/cm3−0.9999 g/cm3) for Δρ , 1.0000 g/cm3 for ρ and (4.0 °C−0 °C) for ΔT in above equation to find β .

β=−(1.0000 g/cm3−0.9999 g/cm3)(1.0000 g/cm3)((4.0 °C−0.0 °C))=−2.5×10−5 °C−1

Conclusion:

Therefore, the value of β for the temperature interval 0 °C to 4.0 °C is −2.5×10−5 (°C)−1 .

Answer 6

Chapter 16, Problem 62P

(a)

To determine

To show: The fraction change in density for a change in temperature is given by Δρρ=−βΔT .

(a)

Expert Solution

Answer to Problem 62P

The fraction change in density for a change in temperature is Δρρ=−βΔT .

Explanation of Solution

Given info: The density of the liquid is ρ , the change in the temperature is ΔT , the maximum density of fresh water at 4.0 °C is 1.0000 g/cm3 , the density of fresh water at 10.0 °C is 0.9997 g/cm3 and the density of water at 0 °C is 0.9999 g/cm3 .

The expression for the coefficient of the volume expansion is,

βΔT=ΔVV

Here,

β is the coefficient of the volume expansion.

The formula for the density is,

ρ=mV

Here,

m is the mass of the water.

V is the volume of the water.\

Differentiate the above equation with respect to volume.

dρ=−mV2dV

For very small change in volume and density the expression is,

Δρ=−mV2ΔV=−mVΔVV

Substitute ρ for mV and βΔT for ΔVV in above equation.

Δρ=−ρβΔTΔρρ=−βΔT

Conclusion:

Therefore, the fraction change in density for a change in temperature is Δρρ=−βΔT .

Answer 7

Chapter 16, Problem 62P

(a)

To determine

To show: The fraction change in density for a change in temperature is given by Δρρ=−βΔT .

Answer 8

(a)

Expert Solution

Answer to Problem 62P

The fraction change in density for a change in temperature is Δρρ=−βΔT .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given info: The density of the liquid is ρ , the change in the temperature is ΔT , the maximum density of fresh water at 4.0 °C is 1.0000 g/cm3 , the density of fresh water at 10.0 °C is 0.9997 g/cm3 and the density of water at 0 °C is 0.9999 g/cm3 .

The expression for the coefficient of the volume expansion is,

βΔT=ΔVV

Here,

β is the coefficient of the volume expansion.

The formula for the density is,

ρ=mV

Here,

m is the mass of the water.

V is the volume of the water.\

Differentiate the above equation with respect to volume.

dρ=−mV2dV

For very small change in volume and density the expression is,

Δρ=−mV2ΔV=−mVΔVV

Substitute ρ for mV and βΔT for ΔVV in above equation.

Δρ=−ρβΔTΔρρ=−βΔT

Conclusion:

Therefore, the fraction change in density for a change in temperature is Δρρ=−βΔT .

Answer 13

(b)

To determine

The significance of the negative sign.

(b)

Expert Solution

Answer to Problem 62P

the negative sign signifies that when the temperature increases, the density decreases.

Explanation of Solution

Given info: The density of the liquid is ρ , the change in the temperature is ΔT , the maximum density of fresh water at 4.0 °C is 1.0000 g/cm3 , the density of fresh water at 10.0 °C is 0.9997 g/cm3 and the density of water at 0 °C is 0.9999 g/cm3 .

The expression for the fraction of the density is,

Δρρ=−βΔT (1)

The negative sign in the above expression shows that as the temperature increases, the density of the water is decreases.

Conclusion:

Therefore, the negative sign signifies that when the temperature increases, the density decreases.

Answer 14

(b)

To determine

The significance of the negative sign.

Answer 15

(b)

Expert Solution

Answer to Problem 62P

the negative sign signifies that when the temperature increases, the density decreases.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Given info: The density of the liquid is ρ , the change in the temperature is ΔT , the maximum density of fresh water at 4.0 °C is 1.0000 g/cm3 , the density of fresh water at 10.0 °C is 0.9997 g/cm3 and the density of water at 0 °C is 0.9999 g/cm3 .

The expression for the fraction of the density is,

Δρρ=−βΔT (1)

The negative sign in the above expression shows that as the temperature increases, the density of the water is decreases.

Conclusion:

Therefore, the negative sign signifies that when the temperature increases, the density decreases.

Answer 20

(c)

To determine

The value of β for the temperature interval 4.0 °C to 10.0 °C .

(c)

Expert Solution

Answer to Problem 62P

The value of β for the temperature interval 4.0 °C to 10.0 °C is 5×10−5 (°C)−1 .

Explanation of Solution

Given info: The density of the liquid is ρ , the change in the temperature is ΔT , the maximum density of fresh water at 4.0 °C is 1.0000 g/cm3 , the density of fresh water at 10.0 °C is 0.9997 g/cm3 and the density of water at 0 °C is 0.9999 g/cm3 .

Recall the equation (1) and rearrange for β .

Δρρ=−βΔTβ=−ΔρρΔT

Substitute (0.9997 g/cm3−1.0000 g/cm3) for Δρ , 1.0000 g/cm3 for ρ and (10.0 °C−4.0 °C) for ΔT in above equation to find β .

β=−(0.9997 g/cm3−1.0000 g/cm3)(1.0000 g/cm3)((10.0 °C−4.0 °C))=5×10−5 °C−1

Conclusion:

Therefore, the value of β for temperature interval 4.0 °C to 10.0 °C is 5×10−5 (°C)−1 .

Answer 21

(c)

To determine

The value of β for the temperature interval 4.0 °C to 10.0 °C .

Answer 22

(c)

Expert Solution

Answer to Problem 62P

The value of β for the temperature interval 4.0 °C to 10.0 °C is 5×10−5 (°C)−1 .

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Given info: The density of the liquid is ρ , the change in the temperature is ΔT , the maximum density of fresh water at 4.0 °C is 1.0000 g/cm3 , the density of fresh water at 10.0 °C is 0.9997 g/cm3 and the density of water at 0 °C is 0.9999 g/cm3 .

Recall the equation (1) and rearrange for β .

Δρρ=−βΔTβ=−ΔρρΔT

Substitute (0.9997 g/cm3−1.0000 g/cm3) for Δρ , 1.0000 g/cm3 for ρ and (10.0 °C−4.0 °C) for ΔT in above equation to find β .

β=−(0.9997 g/cm3−1.0000 g/cm3)(1.0000 g/cm3)((10.0 °C−4.0 °C))=5×10−5 °C−1

Conclusion:

Therefore, the value of β for temperature interval 4.0 °C to 10.0 °C is 5×10−5 (°C)−1 .

Answer 27

(d)

To determine

The value of β for the temperature interval 0 °C to 4.0 °C .

(d)

Expert Solution

Answer to Problem 62P

The value of β for temperature interval 0 °C to 4.0 °C is −2.5×10−5 (°C)−1 .

Explanation of Solution

Given info: The density of the liquid is ρ , the change in the temperature is ΔT , the maximum density of fresh water at 4.0 °C is 1.0000 g/cm3 , the density of fresh water at 10.0 °C is 0.9997 g/cm3 and the density of water at 0 °C is 0.9999 g/cm3 .

The expression for β is,

Δρρ=−βΔTβ=−ΔρρΔT

Substitute (1.0000 g/cm3−0.9999 g/cm3) for Δρ , 1.0000 g/cm3 for ρ and (4.0 °C−0 °C) for ΔT in above equation to find β .

β=−(1.0000 g/cm3−0.9999 g/cm3)(1.0000 g/cm3)((4.0 °C−0.0 °C))=−2.5×10−5 °C−1

Conclusion:

Therefore, the value of β for the temperature interval 0 °C to 4.0 °C is −2.5×10−5 (°C)−1 .

Answer 28

(d)

To determine

The value of β for the temperature interval 0 °C to 4.0 °C .

Answer 29

(d)

Expert Solution

Answer to Problem 62P

The value of β for temperature interval 0 °C to 4.0 °C is −2.5×10−5 (°C)−1 .

Answer 30

(d)

Expert Solution

Answer 31

(d)

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

Given info: The density of the liquid is ρ , the change in the temperature is ΔT , the maximum density of fresh water at 4.0 °C is 1.0000 g/cm3 , the density of fresh water at 10.0 °C is 0.9997 g/cm3 and the density of water at 0 °C is 0.9999 g/cm3 .