The expression for the compressive stress and strain.

Question

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Answer 1

Question

Chapter 11, Problem 14P

(a)

To determine

The expression for the compressive stress and strain.

(a)

Expert Solution

Answer to Problem 14P

The expression for the compressive stress and strain is

$F/A=Y(α(ΔT))$ .

Explanation of Solution

Given Info:

Expression for the compressive stress and strain is,

$FA=Y(ΔLL0)$

F is the force
A is the area,
Y is the Young’s modulus,
$ΔL$ is change in length,
$L0$ is the original length,

Expression for change in length due to linear expansion is,

$ΔLL0=α(ΔT)$

$α$ is the coefficient of linear expansion,
$ΔT$ is the change in temperature,

Use $α(ΔT)$ for $ΛL/L0$ in $F/A=Y(ΔL/L0)$ to rewrite.

$FA=Y(α(ΔT))$

Conclusion:

Thus, the expression for stress is $F/A=Y(α(ΔT))$ .

(b)

To determine

Obtain symbolic equation for thermal energy transfer.

(b)

Expert Solution

Answer to Problem 14P

Symbolic equation for thermal energy transfer is

$Q=(mc/Yα)(F/A)$ .

Explanation of Solution

Given Info

Formula for heat transfer is,

$Q=mcΔT$

Q is the heat transfer,
m is the mass,
c is the specific heat,
$ΔT$ is the change in temperature,

Substitute $F/AYα$ for $ΔT$ (from (a)) to rewrite Q.

$Q=mc(F/AYα)=mcYα(FA)$

Conclusion:

Symbolic equation for thermal energy transfer is $Q=(mc/Yα)(F/A)$ .

(c)

To determine

The mass of the concrete slab.

(c)

Expert Solution

Answer to Problem 14P

The mass of the concrete slab is

$96.0 kg$ .

Explanation of Solution

Given info: Thickness of the slab is 4.00 cm, length of the slab is 1.00 m, and width of the slab is 1.00m and density of the slab is $2.40×103 kg⋅m−3$ .

Formula to calculate the volume of the slab is,

$V=l⋅b⋅w$

V is the volume of the slab,
l is the length of the slab,
b is the breadth of the slab,
w is the thickness of the slab,

Formula to calculate the mass of the concrete slab is,

$m=ρV$

m is the mass of the slab,
$ρ$ is the density of the slab,

Use $l⋅b⋅w$ for V in the above expression to rewrite m.

$m=ρ(l⋅b⋅w)$

Substitute $2.40×103 kg⋅m−3$ for $ρ$ , 1.00 m for l, 1.00 m for b, and 4.00 cm for w to find m.

$m=(2.40×103 kg⋅m−3)[(1.00 m)(1.00 m)(4.00 cm)(1 m102 cm)]=96 kg$

Conclusion:

Therefore, the mass of the concrete slab is $96.0 kg$ .

(d)

To determine

How much thermal energy must be transferred to the slab to reach the compressive stress.

(d)

Expert Solution

Answer to Problem 14P

The thermal energy transferred is

$6.7×106 J$ .

Explanation of Solution

Give info: Ultimate compressive strength of concrete is $2.00×107 Pa$ , specific heat is $880J/kg⋅°C$ , and Young’s modulus is $2.1×1010 Pa$ .

Formula to calculate the thermal energy transfer is,

$QMAX=mcYα(FA)MAX$

Substitute 96.0 kg for m, $2.1×1010 Pa$ for Y, $880J/kg⋅°C$ for c, $2.00×107 Pa$ for $F/A$ , $12×10−6(°C)−1$ for $α$ to find $QMAX$ .

$QMAX=(96.0 kg)(880J/kg⋅°C)(2.1×1010 Pa)(12×10−6(°C)−1)(2.00×107 Pa)=6.7×106 J$

Conclusion:

The maximum thermal energy the slab can absorb before starting to break is $6.7×106 J$ .

(e)

To determine

The change in temperature.

(e)

Expert Solution

Answer to Problem 14P

The change in temperature of the slab is

$79°C$ .

Explanation of Solution

Given info: Maximum thermal energy is $6.7×106 J$ , mass is 96.0 kg, and specific heat of the slab is $880J/kg⋅°C$ .

Formula to calculate change in temperature is,

$ΔT=Qmc$

$ΔT$ is the change in temperature,

Substitute $6.7×106 J$ for Q, 96.0 kg for Q, and $880J/kg⋅°C$ for c to calculate $ΔT$ .

$ΔT=6.7×106 J(96.0 kg)(880J/kg⋅°C)=79.3°C≈79°C$

Conclusion:

The change in temperature of the slab is $79°C$ .

(f)

To determine

The time required for the slab to reach the point of danger of cracking due to compressive stress.

(f)

Expert Solution

Answer to Problem 14P

The time required is

$3.7 h$ .

Explanation of Solution

Given info: Power delivered by the sun is $1.00×103 W$ .

Formula to calculate the time is,

$t=QMAXPabsorb$

t is the time required for the slab to reach the point of cracking,
$Pabsorb$ is the power absorbed by the slab,

On an average half of the energy delivered by the sun in absorbed $Pabsorb=12Psolar$ .

Use $1/2Psolar$ for $Pabsorb$ in the above expression to rewrite t.

$t=QMAX(1/2)Pabsorb$

Substitute $1.00×103 W$ for $Pabsorb$ and $6.7×106 J$ for $QMAX$ to find t.

$t=6.7×106 J(1/2)(1.00×103 W)=13400 s(1 hr3600 s)=3.7 h$

Conclusion:

The time required is $3.7 h$ .

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Answer 2

Question

Chapter 11, Problem 14P

(a)

To determine

The expression for the compressive stress and strain.

(a)

Expert Solution

Answer to Problem 14P

The expression for the compressive stress and strain is

$F/A=Y(α(ΔT))$ .

Explanation of Solution

Given Info:

Expression for the compressive stress and strain is,

$FA=Y(ΔLL0)$

F is the force
A is the area,
Y is the Young’s modulus,
$ΔL$ is change in length,
$L0$ is the original length,

Expression for change in length due to linear expansion is,

$ΔLL0=α(ΔT)$

$α$ is the coefficient of linear expansion,
$ΔT$ is the change in temperature,

Use $α(ΔT)$ for $ΛL/L0$ in $F/A=Y(ΔL/L0)$ to rewrite.

$FA=Y(α(ΔT))$

Conclusion:

Thus, the expression for stress is $F/A=Y(α(ΔT))$ .

(b)

To determine

Obtain symbolic equation for thermal energy transfer.

(b)

Expert Solution

Answer to Problem 14P

Symbolic equation for thermal energy transfer is

$Q=(mc/Yα)(F/A)$ .

Explanation of Solution

Given Info

Formula for heat transfer is,

$Q=mcΔT$

Q is the heat transfer,
m is the mass,
c is the specific heat,
$ΔT$ is the change in temperature,

Substitute $F/AYα$ for $ΔT$ (from (a)) to rewrite Q.

$Q=mc(F/AYα)=mcYα(FA)$

Conclusion:

Symbolic equation for thermal energy transfer is $Q=(mc/Yα)(F/A)$ .

(c)

To determine

The mass of the concrete slab.

(c)

Expert Solution

Answer to Problem 14P

The mass of the concrete slab is

$96.0 kg$ .

Explanation of Solution

Given info: Thickness of the slab is 4.00 cm, length of the slab is 1.00 m, and width of the slab is 1.00m and density of the slab is $2.40×103 kg⋅m−3$ .

Formula to calculate the volume of the slab is,

$V=l⋅b⋅w$

V is the volume of the slab,
l is the length of the slab,
b is the breadth of the slab,
w is the thickness of the slab,

Formula to calculate the mass of the concrete slab is,

$m=ρV$

m is the mass of the slab,
$ρ$ is the density of the slab,

Use $l⋅b⋅w$ for V in the above expression to rewrite m.

$m=ρ(l⋅b⋅w)$

Substitute $2.40×103 kg⋅m−3$ for $ρ$ , 1.00 m for l, 1.00 m for b, and 4.00 cm for w to find m.

$m=(2.40×103 kg⋅m−3)[(1.00 m)(1.00 m)(4.00 cm)(1 m102 cm)]=96 kg$

Conclusion:

Therefore, the mass of the concrete slab is $96.0 kg$ .

(d)

To determine

How much thermal energy must be transferred to the slab to reach the compressive stress.

(d)

Expert Solution

Answer to Problem 14P

The thermal energy transferred is

$6.7×106 J$ .

Explanation of Solution

Give info: Ultimate compressive strength of concrete is $2.00×107 Pa$ , specific heat is $880J/kg⋅°C$ , and Young’s modulus is $2.1×1010 Pa$ .

Formula to calculate the thermal energy transfer is,

$QMAX=mcYα(FA)MAX$

Substitute 96.0 kg for m, $2.1×1010 Pa$ for Y, $880J/kg⋅°C$ for c, $2.00×107 Pa$ for $F/A$ , $12×10−6(°C)−1$ for $α$ to find $QMAX$ .

$QMAX=(96.0 kg)(880J/kg⋅°C)(2.1×1010 Pa)(12×10−6(°C)−1)(2.00×107 Pa)=6.7×106 J$

Conclusion:

The maximum thermal energy the slab can absorb before starting to break is $6.7×106 J$ .

(e)

To determine

The change in temperature.

(e)

Expert Solution

Answer to Problem 14P

The change in temperature of the slab is

$79°C$ .

Explanation of Solution

Given info: Maximum thermal energy is $6.7×106 J$ , mass is 96.0 kg, and specific heat of the slab is $880J/kg⋅°C$ .

Formula to calculate change in temperature is,

$ΔT=Qmc$

$ΔT$ is the change in temperature,

Substitute $6.7×106 J$ for Q, 96.0 kg for Q, and $880J/kg⋅°C$ for c to calculate $ΔT$ .

$ΔT=6.7×106 J(96.0 kg)(880J/kg⋅°C)=79.3°C≈79°C$

Conclusion:

The change in temperature of the slab is $79°C$ .

(f)

To determine

The time required for the slab to reach the point of danger of cracking due to compressive stress.

(f)

Expert Solution

Answer to Problem 14P

The time required is

$3.7 h$ .

Explanation of Solution

Given info: Power delivered by the sun is $1.00×103 W$ .

Formula to calculate the time is,

$t=QMAXPabsorb$

t is the time required for the slab to reach the point of cracking,
$Pabsorb$ is the power absorbed by the slab,

On an average half of the energy delivered by the sun in absorbed $Pabsorb=12Psolar$ .

Use $1/2Psolar$ for $Pabsorb$ in the above expression to rewrite t.

$t=QMAX(1/2)Pabsorb$

Substitute $1.00×103 W$ for $Pabsorb$ and $6.7×106 J$ for $QMAX$ to find t.

$t=6.7×106 J(1/2)(1.00×103 W)=13400 s(1 hr3600 s)=3.7 h$

Conclusion:

The time required is $3.7 h$ .

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Answer 3

Question

Chapter 11, Problem 14P

(a)

To determine

The expression for the compressive stress and strain.

(a)

Expert Solution

Answer to Problem 14P

The expression for the compressive stress and strain is

$F/A=Y(α(ΔT))$ .

Explanation of Solution

Given Info:

Expression for the compressive stress and strain is,

$FA=Y(ΔLL0)$

F is the force
A is the area,
Y is the Young’s modulus,
$ΔL$ is change in length,
$L0$ is the original length,

Expression for change in length due to linear expansion is,

$ΔLL0=α(ΔT)$

$α$ is the coefficient of linear expansion,
$ΔT$ is the change in temperature,

Use $α(ΔT)$ for $ΛL/L0$ in $F/A=Y(ΔL/L0)$ to rewrite.

$FA=Y(α(ΔT))$

Conclusion:

Thus, the expression for stress is $F/A=Y(α(ΔT))$ .

(b)

To determine

Obtain symbolic equation for thermal energy transfer.

(b)

Expert Solution

Answer to Problem 14P

Symbolic equation for thermal energy transfer is

$Q=(mc/Yα)(F/A)$ .

Explanation of Solution

Given Info

Formula for heat transfer is,

$Q=mcΔT$

Q is the heat transfer,
m is the mass,
c is the specific heat,
$ΔT$ is the change in temperature,

Substitute $F/AYα$ for $ΔT$ (from (a)) to rewrite Q.

$Q=mc(F/AYα)=mcYα(FA)$

Conclusion:

Symbolic equation for thermal energy transfer is $Q=(mc/Yα)(F/A)$ .

(c)

To determine

The mass of the concrete slab.

(c)

Expert Solution

Answer to Problem 14P

The mass of the concrete slab is

$96.0 kg$ .

Explanation of Solution

Given info: Thickness of the slab is 4.00 cm, length of the slab is 1.00 m, and width of the slab is 1.00m and density of the slab is $2.40×103 kg⋅m−3$ .

Formula to calculate the volume of the slab is,

$V=l⋅b⋅w$

V is the volume of the slab,
l is the length of the slab,
b is the breadth of the slab,
w is the thickness of the slab,

Formula to calculate the mass of the concrete slab is,

$m=ρV$

m is the mass of the slab,
$ρ$ is the density of the slab,

Use $l⋅b⋅w$ for V in the above expression to rewrite m.

$m=ρ(l⋅b⋅w)$

Substitute $2.40×103 kg⋅m−3$ for $ρ$ , 1.00 m for l, 1.00 m for b, and 4.00 cm for w to find m.

$m=(2.40×103 kg⋅m−3)[(1.00 m)(1.00 m)(4.00 cm)(1 m102 cm)]=96 kg$

Conclusion:

Therefore, the mass of the concrete slab is $96.0 kg$ .

(d)

To determine

How much thermal energy must be transferred to the slab to reach the compressive stress.

(d)

Expert Solution

Answer to Problem 14P

The thermal energy transferred is

$6.7×106 J$ .

Explanation of Solution

Give info: Ultimate compressive strength of concrete is $2.00×107 Pa$ , specific heat is $880J/kg⋅°C$ , and Young’s modulus is $2.1×1010 Pa$ .

Formula to calculate the thermal energy transfer is,

$QMAX=mcYα(FA)MAX$

Substitute 96.0 kg for m, $2.1×1010 Pa$ for Y, $880J/kg⋅°C$ for c, $2.00×107 Pa$ for $F/A$ , $12×10−6(°C)−1$ for $α$ to find $QMAX$ .

$QMAX=(96.0 kg)(880J/kg⋅°C)(2.1×1010 Pa)(12×10−6(°C)−1)(2.00×107 Pa)=6.7×106 J$

Conclusion:

The maximum thermal energy the slab can absorb before starting to break is $6.7×106 J$ .

(e)

To determine

The change in temperature.

(e)

Expert Solution

Answer to Problem 14P

The change in temperature of the slab is

$79°C$ .

Explanation of Solution

Given info: Maximum thermal energy is $6.7×106 J$ , mass is 96.0 kg, and specific heat of the slab is $880J/kg⋅°C$ .

Formula to calculate change in temperature is,

$ΔT=Qmc$

$ΔT$ is the change in temperature,

Substitute $6.7×106 J$ for Q, 96.0 kg for Q, and $880J/kg⋅°C$ for c to calculate $ΔT$ .

$ΔT=6.7×106 J(96.0 kg)(880J/kg⋅°C)=79.3°C≈79°C$

Conclusion:

The change in temperature of the slab is $79°C$ .

(f)

To determine

The time required for the slab to reach the point of danger of cracking due to compressive stress.

(f)

Expert Solution

Answer to Problem 14P

The time required is

$3.7 h$ .

Explanation of Solution

Given info: Power delivered by the sun is $1.00×103 W$ .

Formula to calculate the time is,

$t=QMAXPabsorb$

t is the time required for the slab to reach the point of cracking,
$Pabsorb$ is the power absorbed by the slab,

On an average half of the energy delivered by the sun in absorbed $Pabsorb=12Psolar$ .

Use $1/2Psolar$ for $Pabsorb$ in the above expression to rewrite t.

$t=QMAX(1/2)Pabsorb$

Substitute $1.00×103 W$ for $Pabsorb$ and $6.7×106 J$ for $QMAX$ to find t.

$t=6.7×106 J(1/2)(1.00×103 W)=13400 s(1 hr3600 s)=3.7 h$

Conclusion:

The time required is $3.7 h$ .

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Answer 4

Question

Chapter 11, Problem 14P

(a)

To determine

The expression for the compressive stress and strain.

(a)

Expert Solution

Answer to Problem 14P

The expression for the compressive stress and strain is

$F/A=Y(α(ΔT))$ .

Explanation of Solution

Given Info:

Expression for the compressive stress and strain is,

$FA=Y(ΔLL0)$

F is the force
A is the area,
Y is the Young’s modulus,
$ΔL$ is change in length,
$L0$ is the original length,

Expression for change in length due to linear expansion is,

$ΔLL0=α(ΔT)$

$α$ is the coefficient of linear expansion,
$ΔT$ is the change in temperature,

Use $α(ΔT)$ for $ΛL/L0$ in $F/A=Y(ΔL/L0)$ to rewrite.

$FA=Y(α(ΔT))$

Conclusion:

Thus, the expression for stress is $F/A=Y(α(ΔT))$ .

(b)

To determine

Obtain symbolic equation for thermal energy transfer.

(b)

Expert Solution

Answer to Problem 14P

Symbolic equation for thermal energy transfer is

$Q=(mc/Yα)(F/A)$ .

Explanation of Solution

Given Info

Formula for heat transfer is,

$Q=mcΔT$

Q is the heat transfer,
m is the mass,
c is the specific heat,
$ΔT$ is the change in temperature,

Substitute $F/AYα$ for $ΔT$ (from (a)) to rewrite Q.

$Q=mc(F/AYα)=mcYα(FA)$

Conclusion:

Symbolic equation for thermal energy transfer is $Q=(mc/Yα)(F/A)$ .

(c)

To determine

The mass of the concrete slab.

(c)

Expert Solution

Answer to Problem 14P

The mass of the concrete slab is

$96.0 kg$ .

Explanation of Solution

Given info: Thickness of the slab is 4.00 cm, length of the slab is 1.00 m, and width of the slab is 1.00m and density of the slab is $2.40×103 kg⋅m−3$ .

Formula to calculate the volume of the slab is,

$V=l⋅b⋅w$

V is the volume of the slab,
l is the length of the slab,
b is the breadth of the slab,
w is the thickness of the slab,

Formula to calculate the mass of the concrete slab is,

$m=ρV$

m is the mass of the slab,
$ρ$ is the density of the slab,

Use $l⋅b⋅w$ for V in the above expression to rewrite m.

$m=ρ(l⋅b⋅w)$

Substitute $2.40×103 kg⋅m−3$ for $ρ$ , 1.00 m for l, 1.00 m for b, and 4.00 cm for w to find m.

$m=(2.40×103 kg⋅m−3)[(1.00 m)(1.00 m)(4.00 cm)(1 m102 cm)]=96 kg$

Conclusion:

Therefore, the mass of the concrete slab is $96.0 kg$ .

(d)

To determine

How much thermal energy must be transferred to the slab to reach the compressive stress.

(d)

Expert Solution

Answer to Problem 14P

The thermal energy transferred is

$6.7×106 J$ .

Explanation of Solution

Give info: Ultimate compressive strength of concrete is $2.00×107 Pa$ , specific heat is $880J/kg⋅°C$ , and Young’s modulus is $2.1×1010 Pa$ .

Formula to calculate the thermal energy transfer is,

$QMAX=mcYα(FA)MAX$

Substitute 96.0 kg for m, $2.1×1010 Pa$ for Y, $880J/kg⋅°C$ for c, $2.00×107 Pa$ for $F/A$ , $12×10−6(°C)−1$ for $α$ to find $QMAX$ .

$QMAX=(96.0 kg)(880J/kg⋅°C)(2.1×1010 Pa)(12×10−6(°C)−1)(2.00×107 Pa)=6.7×106 J$

Conclusion:

The maximum thermal energy the slab can absorb before starting to break is $6.7×106 J$ .

(e)

To determine

The change in temperature.

(e)

Expert Solution

Answer to Problem 14P

The change in temperature of the slab is

$79°C$ .

Explanation of Solution

Given info: Maximum thermal energy is $6.7×106 J$ , mass is 96.0 kg, and specific heat of the slab is $880J/kg⋅°C$ .

Formula to calculate change in temperature is,

$ΔT=Qmc$

$ΔT$ is the change in temperature,

Substitute $6.7×106 J$ for Q, 96.0 kg for Q, and $880J/kg⋅°C$ for c to calculate $ΔT$ .

$ΔT=6.7×106 J(96.0 kg)(880J/kg⋅°C)=79.3°C≈79°C$

Conclusion:

The change in temperature of the slab is $79°C$ .

(f)

To determine

The time required for the slab to reach the point of danger of cracking due to compressive stress.

(f)

Expert Solution

Answer to Problem 14P

The time required is

$3.7 h$ .

Explanation of Solution

Given info: Power delivered by the sun is $1.00×103 W$ .

Formula to calculate the time is,

$t=QMAXPabsorb$

t is the time required for the slab to reach the point of cracking,
$Pabsorb$ is the power absorbed by the slab,

On an average half of the energy delivered by the sun in absorbed $Pabsorb=12Psolar$ .

Use $1/2Psolar$ for $Pabsorb$ in the above expression to rewrite t.

$t=QMAX(1/2)Pabsorb$

Substitute $1.00×103 W$ for $Pabsorb$ and $6.7×106 J$ for $QMAX$ to find t.

$t=6.7×106 J(1/2)(1.00×103 W)=13400 s(1 hr3600 s)=3.7 h$

Conclusion:

The time required is $3.7 h$ .

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Answer 5

Chapter 11, Problem 14P

(a)

To determine

The expression for the compressive stress and strain.

(a)

Expert Solution

Answer to Problem 14P

The expression for the compressive stress and strain is

$F/A=Y(α(ΔT))$ .

Explanation of Solution

Given Info:

Expression for the compressive stress and strain is,

$FA=Y(ΔLL0)$

F is the force
A is the area,
Y is the Young’s modulus,
$ΔL$ is change in length,
$L0$ is the original length,

Expression for change in length due to linear expansion is,

$ΔLL0=α(ΔT)$

$α$ is the coefficient of linear expansion,
$ΔT$ is the change in temperature,

Use $α(ΔT)$ for $ΛL/L0$ in $F/A=Y(ΔL/L0)$ to rewrite.

$FA=Y(α(ΔT))$

Conclusion:

Thus, the expression for stress is $F/A=Y(α(ΔT))$ .

(b)

To determine

Obtain symbolic equation for thermal energy transfer.

(b)

Expert Solution

Answer to Problem 14P

Symbolic equation for thermal energy transfer is

$Q=(mc/Yα)(F/A)$ .

Explanation of Solution

Given Info

Formula for heat transfer is,

$Q=mcΔT$

Q is the heat transfer,
m is the mass,
c is the specific heat,
$ΔT$ is the change in temperature,

Substitute $F/AYα$ for $ΔT$ (from (a)) to rewrite Q.

$Q=mc(F/AYα)=mcYα(FA)$

Conclusion:

Symbolic equation for thermal energy transfer is $Q=(mc/Yα)(F/A)$ .

(c)

To determine

The mass of the concrete slab.

(c)

Expert Solution

Answer to Problem 14P

The mass of the concrete slab is

$96.0 kg$ .

Explanation of Solution

Given info: Thickness of the slab is 4.00 cm, length of the slab is 1.00 m, and width of the slab is 1.00m and density of the slab is $2.40×103 kg⋅m−3$ .

Formula to calculate the volume of the slab is,

$V=l⋅b⋅w$

V is the volume of the slab,
l is the length of the slab,
b is the breadth of the slab,
w is the thickness of the slab,

Formula to calculate the mass of the concrete slab is,

$m=ρV$

m is the mass of the slab,
$ρ$ is the density of the slab,

Use $l⋅b⋅w$ for V in the above expression to rewrite m.

$m=ρ(l⋅b⋅w)$

Substitute $2.40×103 kg⋅m−3$ for $ρ$ , 1.00 m for l, 1.00 m for b, and 4.00 cm for w to find m.

$m=(2.40×103 kg⋅m−3)[(1.00 m)(1.00 m)(4.00 cm)(1 m102 cm)]=96 kg$

Conclusion:

Therefore, the mass of the concrete slab is $96.0 kg$ .

(d)

To determine

How much thermal energy must be transferred to the slab to reach the compressive stress.

(d)

Expert Solution

Answer to Problem 14P

The thermal energy transferred is

$6.7×106 J$ .

Explanation of Solution

Give info: Ultimate compressive strength of concrete is $2.00×107 Pa$ , specific heat is $880J/kg⋅°C$ , and Young’s modulus is $2.1×1010 Pa$ .

Formula to calculate the thermal energy transfer is,

$QMAX=mcYα(FA)MAX$

Substitute 96.0 kg for m, $2.1×1010 Pa$ for Y, $880J/kg⋅°C$ for c, $2.00×107 Pa$ for $F/A$ , $12×10−6(°C)−1$ for $α$ to find $QMAX$ .

$QMAX=(96.0 kg)(880J/kg⋅°C)(2.1×1010 Pa)(12×10−6(°C)−1)(2.00×107 Pa)=6.7×106 J$

Conclusion:

The maximum thermal energy the slab can absorb before starting to break is $6.7×106 J$ .

(e)

To determine

The change in temperature.

(e)

Expert Solution

Answer to Problem 14P

The change in temperature of the slab is

$79°C$ .

Explanation of Solution

Given info: Maximum thermal energy is $6.7×106 J$ , mass is 96.0 kg, and specific heat of the slab is $880J/kg⋅°C$ .

Formula to calculate change in temperature is,

$ΔT=Qmc$

$ΔT$ is the change in temperature,

Substitute $6.7×106 J$ for Q, 96.0 kg for Q, and $880J/kg⋅°C$ for c to calculate $ΔT$ .

$ΔT=6.7×106 J(96.0 kg)(880J/kg⋅°C)=79.3°C≈79°C$

Conclusion:

The change in temperature of the slab is $79°C$ .

(f)

To determine

The time required for the slab to reach the point of danger of cracking due to compressive stress.

(f)

Expert Solution

Answer to Problem 14P

The time required is

$3.7 h$ .

Explanation of Solution

Given info: Power delivered by the sun is $1.00×103 W$ .

Formula to calculate the time is,

$t=QMAXPabsorb$

t is the time required for the slab to reach the point of cracking,
$Pabsorb$ is the power absorbed by the slab,

On an average half of the energy delivered by the sun in absorbed $Pabsorb=12Psolar$ .

Use $1/2Psolar$ for $Pabsorb$ in the above expression to rewrite t.

$t=QMAX(1/2)Pabsorb$

Substitute $1.00×103 W$ for $Pabsorb$ and $6.7×106 J$ for $QMAX$ to find t.

$t=6.7×106 J(1/2)(1.00×103 W)=13400 s(1 hr3600 s)=3.7 h$

Conclusion:

The time required is $3.7 h$ .

Answer 6

Chapter 11, Problem 14P

(a)

To determine

The expression for the compressive stress and strain.

(a)

Expert Solution

Answer to Problem 14P

The expression for the compressive stress and strain is

$F/A=Y(α(ΔT))$ .

Explanation of Solution

Given Info:

Expression for the compressive stress and strain is,

$FA=Y(ΔLL0)$

F is the force
A is the area,
Y is the Young’s modulus,
$ΔL$ is change in length,
$L0$ is the original length,

Expression for change in length due to linear expansion is,

$ΔLL0=α(ΔT)$

$α$ is the coefficient of linear expansion,
$ΔT$ is the change in temperature,

Use $α(ΔT)$ for $ΛL/L0$ in $F/A=Y(ΔL/L0)$ to rewrite.

$FA=Y(α(ΔT))$

Conclusion:

Thus, the expression for stress is $F/A=Y(α(ΔT))$ .

Answer 7

Chapter 11, Problem 14P

(a)

To determine

The expression for the compressive stress and strain.

Answer 8

(a)

Expert Solution

Answer to Problem 14P

The expression for the compressive stress and strain is

$F/A=Y(α(ΔT))$ .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given Info:

Expression for the compressive stress and strain is,

$FA=Y(ΔLL0)$

F is the force
A is the area,
Y is the Young’s modulus,
$ΔL$ is change in length,
$L0$ is the original length,

Expression for change in length due to linear expansion is,

$ΔLL0=α(ΔT)$

$α$ is the coefficient of linear expansion,
$ΔT$ is the change in temperature,

Use $α(ΔT)$ for $ΛL/L0$ in $F/A=Y(ΔL/L0)$ to rewrite.

$FA=Y(α(ΔT))$

Conclusion:

Thus, the expression for stress is $F/A=Y(α(ΔT))$ .

Answer 13

(b)

To determine

Obtain symbolic equation for thermal energy transfer.

(b)

Expert Solution

Answer to Problem 14P

Symbolic equation for thermal energy transfer is

$Q=(mc/Yα)(F/A)$ .

Explanation of Solution

Given Info

Formula for heat transfer is,

$Q=mcΔT$

Q is the heat transfer,
m is the mass,
c is the specific heat,
$ΔT$ is the change in temperature,

Substitute $F/AYα$ for $ΔT$ (from (a)) to rewrite Q.

$Q=mc(F/AYα)=mcYα(FA)$

Conclusion:

Symbolic equation for thermal energy transfer is $Q=(mc/Yα)(F/A)$ .

Answer 14

(b)

To determine

Obtain symbolic equation for thermal energy transfer.

Answer 15

(b)

Expert Solution

Answer to Problem 14P

Symbolic equation for thermal energy transfer is

$Q=(mc/Yα)(F/A)$ .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Given Info

Formula for heat transfer is,

$Q=mcΔT$

Q is the heat transfer,
m is the mass,
c is the specific heat,
$ΔT$ is the change in temperature,

Substitute $F/AYα$ for $ΔT$ (from (a)) to rewrite Q.

$Q=mc(F/AYα)=mcYα(FA)$

Conclusion:

Symbolic equation for thermal energy transfer is $Q=(mc/Yα)(F/A)$ .

Answer 20

(c)

To determine

The mass of the concrete slab.

(c)

Expert Solution

Answer to Problem 14P

The mass of the concrete slab is

$96.0 kg$ .

Explanation of Solution

Given info: Thickness of the slab is 4.00 cm, length of the slab is 1.00 m, and width of the slab is 1.00m and density of the slab is $2.40×103 kg⋅m−3$ .

Formula to calculate the volume of the slab is,

$V=l⋅b⋅w$

V is the volume of the slab,
l is the length of the slab,
b is the breadth of the slab,
w is the thickness of the slab,

Formula to calculate the mass of the concrete slab is,

$m=ρV$

m is the mass of the slab,
$ρ$ is the density of the slab,

Use $l⋅b⋅w$ for V in the above expression to rewrite m.

$m=ρ(l⋅b⋅w)$

Substitute $2.40×103 kg⋅m−3$ for $ρ$ , 1.00 m for l, 1.00 m for b, and 4.00 cm for w to find m.

$m=(2.40×103 kg⋅m−3)[(1.00 m)(1.00 m)(4.00 cm)(1 m102 cm)]=96 kg$

Conclusion:

Therefore, the mass of the concrete slab is $96.0 kg$ .

Answer 21

(c)

To determine

The mass of the concrete slab.

Answer 22

(c)

Expert Solution

Answer to Problem 14P

The mass of the concrete slab is

$96.0 kg$ .

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Given info: Thickness of the slab is 4.00 cm, length of the slab is 1.00 m, and width of the slab is 1.00m and density of the slab is $2.40×103 kg⋅m−3$ .

Formula to calculate the volume of the slab is,

$V=l⋅b⋅w$

V is the volume of the slab,
l is the length of the slab,
b is the breadth of the slab,
w is the thickness of the slab,

Formula to calculate the mass of the concrete slab is,

$m=ρV$

m is the mass of the slab,
$ρ$ is the density of the slab,

Use $l⋅b⋅w$ for V in the above expression to rewrite m.

$m=ρ(l⋅b⋅w)$

Substitute $2.40×103 kg⋅m−3$ for $ρ$ , 1.00 m for l, 1.00 m for b, and 4.00 cm for w to find m.

$m=(2.40×103 kg⋅m−3)[(1.00 m)(1.00 m)(4.00 cm)(1 m102 cm)]=96 kg$

Conclusion:

Therefore, the mass of the concrete slab is $96.0 kg$ .

Answer 27

(d)

To determine

How much thermal energy must be transferred to the slab to reach the compressive stress.

(d)

Expert Solution

Answer to Problem 14P

The thermal energy transferred is

$6.7×106 J$ .

Explanation of Solution

Give info: Ultimate compressive strength of concrete is $2.00×107 Pa$ , specific heat is $880J/kg⋅°C$ , and Young’s modulus is $2.1×1010 Pa$ .

Formula to calculate the thermal energy transfer is,

$QMAX=mcYα(FA)MAX$

Substitute 96.0 kg for m, $2.1×1010 Pa$ for Y, $880J/kg⋅°C$ for c, $2.00×107 Pa$ for $F/A$ , $12×10−6(°C)−1$ for $α$ to find $QMAX$ .

$QMAX=(96.0 kg)(880J/kg⋅°C)(2.1×1010 Pa)(12×10−6(°C)−1)(2.00×107 Pa)=6.7×106 J$

Conclusion:

The maximum thermal energy the slab can absorb before starting to break is $6.7×106 J$ .

Answer 28

(d)

To determine

How much thermal energy must be transferred to the slab to reach the compressive stress.

Answer 29

(d)

Expert Solution

Answer to Problem 14P

The thermal energy transferred is

$6.7×106 J$ .

Answer 30

(d)

Expert Solution

Answer 31

(d)

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

Give info: Ultimate compressive strength of concrete is $2.00×107 Pa$ , specific heat is $880J/kg⋅°C$ , and Young’s modulus is $2.1×1010 Pa$ .

Formula to calculate the thermal energy transfer is,

$QMAX=mcYα(FA)MAX$

Substitute 96.0 kg for m, $2.1×1010 Pa$ for Y, $880J/kg⋅°C$ for c, $2.00×107 Pa$ for $F/A$ , $12×10−6(°C)−1$ for $α$ to find $QMAX$ .

$QMAX=(96.0 kg)(880J/kg⋅°C)(2.1×1010 Pa)(12×10−6(°C)−1)(2.00×107 Pa)=6.7×106 J$

Conclusion:

The maximum thermal energy the slab can absorb before starting to break is $6.7×106 J$ .

Answer 34

(e)

To determine

The change in temperature.

(e)

Expert Solution

Answer to Problem 14P

The change in temperature of the slab is

$79°C$ .

Explanation of Solution

Given info: Maximum thermal energy is $6.7×106 J$ , mass is 96.0 kg, and specific heat of the slab is $880J/kg⋅°C$ .

Formula to calculate change in temperature is,

$ΔT=Qmc$

$ΔT$ is the change in temperature,

Substitute $6.7×106 J$ for Q, 96.0 kg for Q, and $880J/kg⋅°C$ for c to calculate $ΔT$ .

$ΔT=6.7×106 J(96.0 kg)(880J/kg⋅°C)=79.3°C≈79°C$

Conclusion:

The change in temperature of the slab is $79°C$ .

Answer 35

(e)

To determine

The change in temperature.

Answer 36

(e)

Expert Solution

Answer to Problem 14P

The change in temperature of the slab is

$79°C$ .

Answer 37

(e)

Expert Solution

Answer 38

(e)

Expert Solution

Answer 39

Expert Solution

Answer 40

Explanation of Solution

Given info: Maximum thermal energy is $6.7×106 J$ , mass is 96.0 kg, and specific heat of the slab is $880J/kg⋅°C$ .

Formula to calculate change in temperature is,

$ΔT=Qmc$

$ΔT$ is the change in temperature,

Substitute $6.7×106 J$ for Q, 96.0 kg for Q, and $880J/kg⋅°C$ for c to calculate $ΔT$ .

$ΔT=6.7×106 J(96.0 kg)(880J/kg⋅°C)=79.3°C≈79°C$

Conclusion:

The change in temperature of the slab is $79°C$ .

Answer 41

(f)

To determine

The time required for the slab to reach the point of danger of cracking due to compressive stress.

(f)

Expert Solution

Answer to Problem 14P

The time required is

$3.7 h$ .

Explanation of Solution

Given info: Power delivered by the sun is $1.00×103 W$ .

Formula to calculate the time is,

$t=QMAXPabsorb$

t is the time required for the slab to reach the point of cracking,
$Pabsorb$ is the power absorbed by the slab,

On an average half of the energy delivered by the sun in absorbed $Pabsorb=12Psolar$ .

Use $1/2Psolar$ for $Pabsorb$ in the above expression to rewrite t.

$t=QMAX(1/2)Pabsorb$

Substitute $1.00×103 W$ for $Pabsorb$ and $6.7×106 J$ for $QMAX$ to find t.

$t=6.7×106 J(1/2)(1.00×103 W)=13400 s(1 hr3600 s)=3.7 h$

Conclusion:

The time required is $3.7 h$ .

Answer 42

(f)

To determine

The time required for the slab to reach the point of danger of cracking due to compressive stress.

Answer 43

(f)

Expert Solution

Answer to Problem 14P

The time required is

$3.7 h$ .

Answer 44

(f)

Expert Solution

Answer 45

(f)

Expert Solution

Answer 46

Expert Solution

Answer 47

Explanation of Solution

Given info: Power delivered by the sun is $1.00×103 W$ .

Formula to calculate the time is,

$t=QMAXPabsorb$

t is the time required for the slab to reach the point of cracking,
$Pabsorb$ is the power absorbed by the slab,

On an average half of the energy delivered by the sun in absorbed $Pabsorb=12Psolar$ .

Use $1/2Psolar$ for $Pabsorb$ in the above expression to rewrite t.

$t=QMAX(1/2)Pabsorb$

Substitute $1.00×103 W$ for $Pabsorb$ and $6.7×106 J$ for $QMAX$ to find t.

$t=6.7×106 J(1/2)(1.00×103 W)=13400 s(1 hr3600 s)=3.7 h$

Conclusion:

The time required is $3.7 h$ .

Answer 48

Ch. 11.2 - Prob. 11.1QQ Ch. 11.4 - Prob. 11.2QQ Ch. 11.5 - Will an ice cube wrapped in a wool blanket remain...Ch. 11.5 - Two rods of the same length and diameter are made...Ch. 11.5 - Stars A and B have the same temperature, but star...Ch. 11 - Prob. 1WUE Ch. 11 - Physics Review An athlete lifts a 175-kg barbell...Ch. 11 - Prob. 3WUE Ch. 11 - Convert 3.50 103 cal to the equivalent number of...Ch. 11 - Prob. 5WUE

The expression for the compressive stress and strain.

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The expression for the compressive stress and strain.

Answer to Problem 14P

Explanation of Solution

Obtain symbolic equation for thermal energy transfer.

Answer to Problem 14P

Explanation of Solution

The mass of the concrete slab.

Answer to Problem 14P

Explanation of Solution

How much thermal energy must be transferred to the slab to reach the compressive stress.

Answer to Problem 14P

Explanation of Solution

The change in temperature.

Answer to Problem 14P

Explanation of Solution

The time required for the slab to reach the point of danger of cracking due to compressive stress.

Answer to Problem 14P

Explanation of Solution

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