Use the coefficient of volume expansion to estimate the density of water as it is heated from 60 ° F to 130 ° F at 1 atm. Compare your result with the actual density (from the appendices).

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Answer 1

Textbook Question

Chapter 2, Problem 39P

Use the coefficient of volume expansion to estimate the density of water as it is heated from $60 ° F$ to $130 ° F$ at 1 atm. Compare your result with the actual density (from the appendices).

Expert Solution & Answer

To determine

The density of water.

Answer to Problem 39P

The density calculated is $986.78 kg/m3$ .

The density calculated and the actual density from appendix is nearly same.

Explanation of Solution

Given information:

The water is heated from $60°F$ to $130°F$ at $1 atm$ .

Write the expression for the average temperature.

$Tavg=T1+T22$ ....... (I)

Here, initial temperature is $T1$ . Final temperature is $T2$ and the average temperature is $Tavg$ .

Write the expression for the density of water.

$Δρ=−ρβΔT$ ....... (II)

Here, the change in density is $Δρ$ , coefficient of volume expansion is $β$ , the temperature difference is $ΔT,$ and the initial density is $ρ$ .

Write the expression for temperature difference.

$ΔT=T2−T1$ ....... (III)

Here, initial temperature is $T1$ . Final temperature is $T2$ and the average temperature is $Tavg$ .

Write the expression for final density.

$Δρ=ρ2−ρρ2=Δρ+ρ$ ....... (IV)

Write the formula for interpolation of two variables.

$ρ2′=(x2−x1)(y3−y1)(x3−x1)+y1$ ....... (V)

Here, the temperature is denoted by variables $x$ , density is denoted by variables $y,$ and the unknown density is $ρ2′$ .

Calculation:

Substitute $60°F$ for $T1$ and $130°F$ for $T2$ in Equation (I).

$Tavg=60°F+130°F2=( 60 °F−32 1.8)°C+( 130 °F−32 1.8)°C2=15.56°C+54.44°C2=35°C$

Refer to Table A-3, "Properties of saturated water" to obtain the value of $β$ as $0.00034 K−1$ at $35°C$ .

Substitute $60°F$ for $T1$ and $130°F$ for $T2$ in Equation (III).

$ΔT=130°F−60°F=(130°F−321.8)°C−(60°F−321.8)°C=(54.44°C+273) K−(15.56°C+273) K=38.88 K$

Substitute $1000 kg/m3$ for $ρ$ , $0.00034 K−1$ for $β$ and $38.88 K$ for $ΔT$ in Equation (II).

$Δρ=−(1000 kg/m3)(0.00034 K−1)(38.88 K)=−(13.219 kg/m3)$

Substitute $1000 kg/m3$ for $ρ$ , $−(13.219 kg/m3)$ for $Δρ$ in Equation (IV).

$ρ2=−13.219 kg/m3+1000 kg/m3=986.78 kg/m3$

Refer to Table $A-3$ , "Properties of saturated water" to obtain the values of $x2$ as $54.44°C$ , $x1$ as $50°C$ , $x3$ as $55°C$ .

Refer to Table $A-3$ , "Properties of saturated water" to obtain the values of $y1$ as $988.1 kg/m3$ , $y3$ as $985.2 kg/m3$ .

Prepare the table for temperature and density of water.

Temperature, $°F$	Density, $kg/m3$
$50 (x1)$	$988.1 (y1)$
$54.44 (x2)$	$? (ρ2′)$
$55 (x3)$	$985.2 (y3)$

Substitute $50°C$ for $x1$ , $54.44°C$ for $x2$ , $55°C$ for $x3$ , $988.1 kg/m3$ for $y1$ and $985.2 kg/m3$ for $y3$ in Equation (V).

$ρ2′=(985.2 kg/ m 3−988.1 kg/ m 3)(54.44°C−50°C)(55°C−50°C)+988.1 kg/m3=((−2.9 kg/ m 3)4.5°C5°C)+988.1 kg/m3=−2.61 kg/m3+988.1 kg/m3=985.49 kg/m3$

Conclusion:

The density calculated is $986.78 kg/m3$ .

The density calculated and the actual density from appendix is nearly same.

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Answer 2

Textbook Question

Chapter 2, Problem 39P

Use the coefficient of volume expansion to estimate the density of water as it is heated from $60 ° F$ to $130 ° F$ at 1 atm. Compare your result with the actual density (from the appendices).

Expert Solution & Answer

To determine

The density of water.

Answer to Problem 39P

The density calculated is $986.78 kg/m3$ .

The density calculated and the actual density from appendix is nearly same.

Explanation of Solution

Given information:

The water is heated from $60°F$ to $130°F$ at $1 atm$ .

Write the expression for the average temperature.

$Tavg=T1+T22$ ....... (I)

Here, initial temperature is $T1$ . Final temperature is $T2$ and the average temperature is $Tavg$ .

Write the expression for the density of water.

$Δρ=−ρβΔT$ ....... (II)

Here, the change in density is $Δρ$ , coefficient of volume expansion is $β$ , the temperature difference is $ΔT,$ and the initial density is $ρ$ .

Write the expression for temperature difference.

$ΔT=T2−T1$ ....... (III)

Here, initial temperature is $T1$ . Final temperature is $T2$ and the average temperature is $Tavg$ .

Write the expression for final density.

$Δρ=ρ2−ρρ2=Δρ+ρ$ ....... (IV)

Write the formula for interpolation of two variables.

$ρ2′=(x2−x1)(y3−y1)(x3−x1)+y1$ ....... (V)

Here, the temperature is denoted by variables $x$ , density is denoted by variables $y,$ and the unknown density is $ρ2′$ .

Calculation:

Substitute $60°F$ for $T1$ and $130°F$ for $T2$ in Equation (I).

$Tavg=60°F+130°F2=( 60 °F−32 1.8)°C+( 130 °F−32 1.8)°C2=15.56°C+54.44°C2=35°C$

Refer to Table A-3, "Properties of saturated water" to obtain the value of $β$ as $0.00034 K−1$ at $35°C$ .

Substitute $60°F$ for $T1$ and $130°F$ for $T2$ in Equation (III).

$ΔT=130°F−60°F=(130°F−321.8)°C−(60°F−321.8)°C=(54.44°C+273) K−(15.56°C+273) K=38.88 K$

Substitute $1000 kg/m3$ for $ρ$ , $0.00034 K−1$ for $β$ and $38.88 K$ for $ΔT$ in Equation (II).

$Δρ=−(1000 kg/m3)(0.00034 K−1)(38.88 K)=−(13.219 kg/m3)$

Substitute $1000 kg/m3$ for $ρ$ , $−(13.219 kg/m3)$ for $Δρ$ in Equation (IV).

$ρ2=−13.219 kg/m3+1000 kg/m3=986.78 kg/m3$

Refer to Table $A-3$ , "Properties of saturated water" to obtain the values of $x2$ as $54.44°C$ , $x1$ as $50°C$ , $x3$ as $55°C$ .

Refer to Table $A-3$ , "Properties of saturated water" to obtain the values of $y1$ as $988.1 kg/m3$ , $y3$ as $985.2 kg/m3$ .

Prepare the table for temperature and density of water.

Temperature, $°F$	Density, $kg/m3$
$50 (x1)$	$988.1 (y1)$
$54.44 (x2)$	$? (ρ2′)$
$55 (x3)$	$985.2 (y3)$

Substitute $50°C$ for $x1$ , $54.44°C$ for $x2$ , $55°C$ for $x3$ , $988.1 kg/m3$ for $y1$ and $985.2 kg/m3$ for $y3$ in Equation (V).

$ρ2′=(985.2 kg/ m 3−988.1 kg/ m 3)(54.44°C−50°C)(55°C−50°C)+988.1 kg/m3=((−2.9 kg/ m 3)4.5°C5°C)+988.1 kg/m3=−2.61 kg/m3+988.1 kg/m3=985.49 kg/m3$

Conclusion:

The density calculated is $986.78 kg/m3$ .

The density calculated and the actual density from appendix is nearly same.

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Answer 3

Textbook Question

Chapter 2, Problem 39P

Use the coefficient of volume expansion to estimate the density of water as it is heated from $60 ° F$ to $130 ° F$ at 1 atm. Compare your result with the actual density (from the appendices).

Expert Solution & Answer

To determine

The density of water.

Answer to Problem 39P

The density calculated is $986.78 kg/m3$ .

The density calculated and the actual density from appendix is nearly same.

Explanation of Solution

Given information:

The water is heated from $60°F$ to $130°F$ at $1 atm$ .

Write the expression for the average temperature.

$Tavg=T1+T22$ ....... (I)

Here, initial temperature is $T1$ . Final temperature is $T2$ and the average temperature is $Tavg$ .

Write the expression for the density of water.

$Δρ=−ρβΔT$ ....... (II)

Here, the change in density is $Δρ$ , coefficient of volume expansion is $β$ , the temperature difference is $ΔT,$ and the initial density is $ρ$ .

Write the expression for temperature difference.

$ΔT=T2−T1$ ....... (III)

Here, initial temperature is $T1$ . Final temperature is $T2$ and the average temperature is $Tavg$ .

Write the expression for final density.

$Δρ=ρ2−ρρ2=Δρ+ρ$ ....... (IV)

Write the formula for interpolation of two variables.

$ρ2′=(x2−x1)(y3−y1)(x3−x1)+y1$ ....... (V)

Here, the temperature is denoted by variables $x$ , density is denoted by variables $y,$ and the unknown density is $ρ2′$ .

Calculation:

Substitute $60°F$ for $T1$ and $130°F$ for $T2$ in Equation (I).

$Tavg=60°F+130°F2=( 60 °F−32 1.8)°C+( 130 °F−32 1.8)°C2=15.56°C+54.44°C2=35°C$

Refer to Table A-3, "Properties of saturated water" to obtain the value of $β$ as $0.00034 K−1$ at $35°C$ .

Substitute $60°F$ for $T1$ and $130°F$ for $T2$ in Equation (III).

$ΔT=130°F−60°F=(130°F−321.8)°C−(60°F−321.8)°C=(54.44°C+273) K−(15.56°C+273) K=38.88 K$

Substitute $1000 kg/m3$ for $ρ$ , $0.00034 K−1$ for $β$ and $38.88 K$ for $ΔT$ in Equation (II).

$Δρ=−(1000 kg/m3)(0.00034 K−1)(38.88 K)=−(13.219 kg/m3)$

Substitute $1000 kg/m3$ for $ρ$ , $−(13.219 kg/m3)$ for $Δρ$ in Equation (IV).

$ρ2=−13.219 kg/m3+1000 kg/m3=986.78 kg/m3$

Refer to Table $A-3$ , "Properties of saturated water" to obtain the values of $x2$ as $54.44°C$ , $x1$ as $50°C$ , $x3$ as $55°C$ .

Refer to Table $A-3$ , "Properties of saturated water" to obtain the values of $y1$ as $988.1 kg/m3$ , $y3$ as $985.2 kg/m3$ .

Prepare the table for temperature and density of water.

Temperature, $°F$	Density, $kg/m3$
$50 (x1)$	$988.1 (y1)$
$54.44 (x2)$	$? (ρ2′)$
$55 (x3)$	$985.2 (y3)$

Substitute $50°C$ for $x1$ , $54.44°C$ for $x2$ , $55°C$ for $x3$ , $988.1 kg/m3$ for $y1$ and $985.2 kg/m3$ for $y3$ in Equation (V).

$ρ2′=(985.2 kg/ m 3−988.1 kg/ m 3)(54.44°C−50°C)(55°C−50°C)+988.1 kg/m3=((−2.9 kg/ m 3)4.5°C5°C)+988.1 kg/m3=−2.61 kg/m3+988.1 kg/m3=985.49 kg/m3$

Conclusion:

The density calculated is $986.78 kg/m3$ .

The density calculated and the actual density from appendix is nearly same.

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Answer 4

Textbook Question

Chapter 2, Problem 39P

Use the coefficient of volume expansion to estimate the density of water as it is heated from $60 ° F$ to $130 ° F$ at 1 atm. Compare your result with the actual density (from the appendices).

Expert Solution & Answer

To determine

The density of water.

Answer to Problem 39P

The density calculated is $986.78 kg/m3$ .

The density calculated and the actual density from appendix is nearly same.

Explanation of Solution

Given information:

The water is heated from $60°F$ to $130°F$ at $1 atm$ .

Write the expression for the average temperature.

$Tavg=T1+T22$ ....... (I)

Here, initial temperature is $T1$ . Final temperature is $T2$ and the average temperature is $Tavg$ .

Write the expression for the density of water.

$Δρ=−ρβΔT$ ....... (II)

Here, the change in density is $Δρ$ , coefficient of volume expansion is $β$ , the temperature difference is $ΔT,$ and the initial density is $ρ$ .

Write the expression for temperature difference.

$ΔT=T2−T1$ ....... (III)

Here, initial temperature is $T1$ . Final temperature is $T2$ and the average temperature is $Tavg$ .

Write the expression for final density.

$Δρ=ρ2−ρρ2=Δρ+ρ$ ....... (IV)

Write the formula for interpolation of two variables.

$ρ2′=(x2−x1)(y3−y1)(x3−x1)+y1$ ....... (V)

Here, the temperature is denoted by variables $x$ , density is denoted by variables $y,$ and the unknown density is $ρ2′$ .

Calculation:

Substitute $60°F$ for $T1$ and $130°F$ for $T2$ in Equation (I).

$Tavg=60°F+130°F2=( 60 °F−32 1.8)°C+( 130 °F−32 1.8)°C2=15.56°C+54.44°C2=35°C$

Refer to Table A-3, "Properties of saturated water" to obtain the value of $β$ as $0.00034 K−1$ at $35°C$ .

Substitute $60°F$ for $T1$ and $130°F$ for $T2$ in Equation (III).

$ΔT=130°F−60°F=(130°F−321.8)°C−(60°F−321.8)°C=(54.44°C+273) K−(15.56°C+273) K=38.88 K$

Substitute $1000 kg/m3$ for $ρ$ , $0.00034 K−1$ for $β$ and $38.88 K$ for $ΔT$ in Equation (II).

$Δρ=−(1000 kg/m3)(0.00034 K−1)(38.88 K)=−(13.219 kg/m3)$

Substitute $1000 kg/m3$ for $ρ$ , $−(13.219 kg/m3)$ for $Δρ$ in Equation (IV).

$ρ2=−13.219 kg/m3+1000 kg/m3=986.78 kg/m3$

Refer to Table $A-3$ , "Properties of saturated water" to obtain the values of $x2$ as $54.44°C$ , $x1$ as $50°C$ , $x3$ as $55°C$ .

Refer to Table $A-3$ , "Properties of saturated water" to obtain the values of $y1$ as $988.1 kg/m3$ , $y3$ as $985.2 kg/m3$ .

Prepare the table for temperature and density of water.

Temperature, $°F$	Density, $kg/m3$
$50 (x1)$	$988.1 (y1)$
$54.44 (x2)$	$? (ρ2′)$
$55 (x3)$	$985.2 (y3)$

Substitute $50°C$ for $x1$ , $54.44°C$ for $x2$ , $55°C$ for $x3$ , $988.1 kg/m3$ for $y1$ and $985.2 kg/m3$ for $y3$ in Equation (V).

$ρ2′=(985.2 kg/ m 3−988.1 kg/ m 3)(54.44°C−50°C)(55°C−50°C)+988.1 kg/m3=((−2.9 kg/ m 3)4.5°C5°C)+988.1 kg/m3=−2.61 kg/m3+988.1 kg/m3=985.49 kg/m3$

Conclusion:

The density calculated is $986.78 kg/m3$ .

The density calculated and the actual density from appendix is nearly same.

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

Expert Solution & Answer

To determine

The density of water.

Answer to Problem 39P

The density calculated is $986.78 kg/m3$ .

The density calculated and the actual density from appendix is nearly same.

Explanation of Solution

Given information:

The water is heated from $60°F$ to $130°F$ at $1 atm$ .

Write the expression for the average temperature.

$Tavg=T1+T22$ ....... (I)

Here, initial temperature is $T1$ . Final temperature is $T2$ and the average temperature is $Tavg$ .

Write the expression for the density of water.

$Δρ=−ρβΔT$ ....... (II)

Here, the change in density is $Δρ$ , coefficient of volume expansion is $β$ , the temperature difference is $ΔT,$ and the initial density is $ρ$ .

Write the expression for temperature difference.

$ΔT=T2−T1$ ....... (III)

Here, initial temperature is $T1$ . Final temperature is $T2$ and the average temperature is $Tavg$ .

Write the expression for final density.

$Δρ=ρ2−ρρ2=Δρ+ρ$ ....... (IV)

Write the formula for interpolation of two variables.

$ρ2′=(x2−x1)(y3−y1)(x3−x1)+y1$ ....... (V)

Here, the temperature is denoted by variables $x$ , density is denoted by variables $y,$ and the unknown density is $ρ2′$ .

Calculation:

Substitute $60°F$ for $T1$ and $130°F$ for $T2$ in Equation (I).

$Tavg=60°F+130°F2=( 60 °F−32 1.8)°C+( 130 °F−32 1.8)°C2=15.56°C+54.44°C2=35°C$

Refer to Table A-3, "Properties of saturated water" to obtain the value of $β$ as $0.00034 K−1$ at $35°C$ .

Substitute $60°F$ for $T1$ and $130°F$ for $T2$ in Equation (III).

$ΔT=130°F−60°F=(130°F−321.8)°C−(60°F−321.8)°C=(54.44°C+273) K−(15.56°C+273) K=38.88 K$

Substitute $1000 kg/m3$ for $ρ$ , $0.00034 K−1$ for $β$ and $38.88 K$ for $ΔT$ in Equation (II).

$Δρ=−(1000 kg/m3)(0.00034 K−1)(38.88 K)=−(13.219 kg/m3)$

Substitute $1000 kg/m3$ for $ρ$ , $−(13.219 kg/m3)$ for $Δρ$ in Equation (IV).

$ρ2=−13.219 kg/m3+1000 kg/m3=986.78 kg/m3$

Refer to Table $A-3$ , "Properties of saturated water" to obtain the values of $x2$ as $54.44°C$ , $x1$ as $50°C$ , $x3$ as $55°C$ .

Refer to Table $A-3$ , "Properties of saturated water" to obtain the values of $y1$ as $988.1 kg/m3$ , $y3$ as $985.2 kg/m3$ .

Prepare the table for temperature and density of water.

Temperature, $°F$	Density, $kg/m3$
$50 (x1)$	$988.1 (y1)$
$54.44 (x2)$	$? (ρ2′)$
$55 (x3)$	$985.2 (y3)$

Substitute $50°C$ for $x1$ , $54.44°C$ for $x2$ , $55°C$ for $x3$ , $988.1 kg/m3$ for $y1$ and $985.2 kg/m3$ for $y3$ in Equation (V).

$ρ2′=(985.2 kg/ m 3−988.1 kg/ m 3)(54.44°C−50°C)(55°C−50°C)+988.1 kg/m3=((−2.9 kg/ m 3)4.5°C5°C)+988.1 kg/m3=−2.61 kg/m3+988.1 kg/m3=985.49 kg/m3$

Conclusion:

The density calculated is $986.78 kg/m3$ .

The density calculated and the actual density from appendix is nearly same.

Answer 8

Expert Solution & Answer

To determine

The density of water.

Answer to Problem 39P

The density calculated is $986.78 kg/m3$ .

The density calculated and the actual density from appendix is nearly same.

Explanation of Solution

Given information:

The water is heated from $60°F$ to $130°F$ at $1 atm$ .

Write the expression for the average temperature.

$Tavg=T1+T22$ ....... (I)

Here, initial temperature is $T1$ . Final temperature is $T2$ and the average temperature is $Tavg$ .

Write the expression for the density of water.

$Δρ=−ρβΔT$ ....... (II)

Here, the change in density is $Δρ$ , coefficient of volume expansion is $β$ , the temperature difference is $ΔT,$ and the initial density is $ρ$ .

Write the expression for temperature difference.

$ΔT=T2−T1$ ....... (III)

Here, initial temperature is $T1$ . Final temperature is $T2$ and the average temperature is $Tavg$ .

Write the expression for final density.

$Δρ=ρ2−ρρ2=Δρ+ρ$ ....... (IV)

Write the formula for interpolation of two variables.

$ρ2′=(x2−x1)(y3−y1)(x3−x1)+y1$ ....... (V)

Here, the temperature is denoted by variables $x$ , density is denoted by variables $y,$ and the unknown density is $ρ2′$ .

Calculation:

Substitute $60°F$ for $T1$ and $130°F$ for $T2$ in Equation (I).

$Tavg=60°F+130°F2=( 60 °F−32 1.8)°C+( 130 °F−32 1.8)°C2=15.56°C+54.44°C2=35°C$

Refer to Table A-3, "Properties of saturated water" to obtain the value of $β$ as $0.00034 K−1$ at $35°C$ .

Substitute $60°F$ for $T1$ and $130°F$ for $T2$ in Equation (III).

$ΔT=130°F−60°F=(130°F−321.8)°C−(60°F−321.8)°C=(54.44°C+273) K−(15.56°C+273) K=38.88 K$

Substitute $1000 kg/m3$ for $ρ$ , $0.00034 K−1$ for $β$ and $38.88 K$ for $ΔT$ in Equation (II).

$Δρ=−(1000 kg/m3)(0.00034 K−1)(38.88 K)=−(13.219 kg/m3)$

Substitute $1000 kg/m3$ for $ρ$ , $−(13.219 kg/m3)$ for $Δρ$ in Equation (IV).

$ρ2=−13.219 kg/m3+1000 kg/m3=986.78 kg/m3$

Refer to Table $A-3$ , "Properties of saturated water" to obtain the values of $x2$ as $54.44°C$ , $x1$ as $50°C$ , $x3$ as $55°C$ .

Refer to Table $A-3$ , "Properties of saturated water" to obtain the values of $y1$ as $988.1 kg/m3$ , $y3$ as $985.2 kg/m3$ .

Prepare the table for temperature and density of water.

Temperature, $°F$	Density, $kg/m3$
$50 (x1)$	$988.1 (y1)$
$54.44 (x2)$	$? (ρ2′)$
$55 (x3)$	$985.2 (y3)$

Substitute $50°C$ for $x1$ , $54.44°C$ for $x2$ , $55°C$ for $x3$ , $988.1 kg/m3$ for $y1$ and $985.2 kg/m3$ for $y3$ in Equation (V).

$ρ2′=(985.2 kg/ m 3−988.1 kg/ m 3)(54.44°C−50°C)(55°C−50°C)+988.1 kg/m3=((−2.9 kg/ m 3)4.5°C5°C)+988.1 kg/m3=−2.61 kg/m3+988.1 kg/m3=985.49 kg/m3$

Conclusion:

The density calculated is $986.78 kg/m3$ .

The density calculated and the actual density from appendix is nearly same.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

To determine

The density of water.

Answer 12

Answer to Problem 39P

The density calculated is $986.78 kg/m3$ .

The density calculated and the actual density from appendix is nearly same.

Answer 13

Explanation of Solution

Given information:

The water is heated from $60°F$ to $130°F$ at $1 atm$ .

Write the expression for the average temperature.

$Tavg=T1+T22$ ....... (I)

Here, initial temperature is $T1$ . Final temperature is $T2$ and the average temperature is $Tavg$ .

Write the expression for the density of water.

$Δρ=−ρβΔT$ ....... (II)

Here, the change in density is $Δρ$ , coefficient of volume expansion is $β$ , the temperature difference is $ΔT,$ and the initial density is $ρ$ .

Write the expression for temperature difference.

$ΔT=T2−T1$ ....... (III)

Here, initial temperature is $T1$ . Final temperature is $T2$ and the average temperature is $Tavg$ .

Write the expression for final density.

$Δρ=ρ2−ρρ2=Δρ+ρ$ ....... (IV)

Write the formula for interpolation of two variables.

$ρ2′=(x2−x1)(y3−y1)(x3−x1)+y1$ ....... (V)

Here, the temperature is denoted by variables $x$ , density is denoted by variables $y,$ and the unknown density is $ρ2′$ .

Calculation:

Substitute $60°F$ for $T1$ and $130°F$ for $T2$ in Equation (I).

$Tavg=60°F+130°F2=( 60 °F−32 1.8)°C+( 130 °F−32 1.8)°C2=15.56°C+54.44°C2=35°C$

Refer to Table A-3, "Properties of saturated water" to obtain the value of $β$ as $0.00034 K−1$ at $35°C$ .

Substitute $60°F$ for $T1$ and $130°F$ for $T2$ in Equation (III).

$ΔT=130°F−60°F=(130°F−321.8)°C−(60°F−321.8)°C=(54.44°C+273) K−(15.56°C+273) K=38.88 K$

Substitute $1000 kg/m3$ for $ρ$ , $0.00034 K−1$ for $β$ and $38.88 K$ for $ΔT$ in Equation (II).

$Δρ=−(1000 kg/m3)(0.00034 K−1)(38.88 K)=−(13.219 kg/m3)$

Substitute $1000 kg/m3$ for $ρ$ , $−(13.219 kg/m3)$ for $Δρ$ in Equation (IV).

$ρ2=−13.219 kg/m3+1000 kg/m3=986.78 kg/m3$

Refer to Table $A-3$ , "Properties of saturated water" to obtain the values of $x2$ as $54.44°C$ , $x1$ as $50°C$ , $x3$ as $55°C$ .

Refer to Table $A-3$ , "Properties of saturated water" to obtain the values of $y1$ as $988.1 kg/m3$ , $y3$ as $985.2 kg/m3$ .

Prepare the table for temperature and density of water.

Temperature, $°F$	Density, $kg/m3$
$50 (x1)$	$988.1 (y1)$
$54.44 (x2)$	$? (ρ2′)$
$55 (x3)$	$985.2 (y3)$

Substitute $50°C$ for $x1$ , $54.44°C$ for $x2$ , $55°C$ for $x3$ , $988.1 kg/m3$ for $y1$ and $985.2 kg/m3$ for $y3$ in Equation (V).

$ρ2′=(985.2 kg/ m 3−988.1 kg/ m 3)(54.44°C−50°C)(55°C−50°C)+988.1 kg/m3=((−2.9 kg/ m 3)4.5°C5°C)+988.1 kg/m3=−2.61 kg/m3+988.1 kg/m3=985.49 kg/m3$

Conclusion:

The density calculated is $986.78 kg/m3$ .

The density calculated and the actual density from appendix is nearly same.

Answer 14

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Use the coefficient of volume expansion to estimate the density of water as it is heated from 60 ° F to 130 ° F at 1 atm. Compare your result with the actual density (from the appendices).

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