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Question: The density of a lead at \[0^\circ {\text{C}}\] is \[{\text{11}}{\text{.34}}\,{\text{gm/cc}}\]. What...

The density of a lead at 0C0^\circ {\text{C}} is 11.34gm/cc{\text{11}}{\text{.34}}\,{\text{gm/cc}}. What is the density of lead at 100C100^\circ {\text{C}}, given the coefficient of linear expansion of lead =28×106/C28 \times {10^{ - 6}}/^\circ {\text{C}}:
A. 13.25g/cm3{\text{13}}{\text{.25}}\,{\text{g/c}}{{\text{m}}^3}
B. 17.25g/cm3{\text{17}}{\text{.25}}\,{\text{g/c}}{{\text{m}}^3}
C. 18.25g/cm3{\text{18}}{\text{.25}}\,{\text{g/c}}{{\text{m}}^3}
D. 11.25g/cm3{\text{11}}{\text{.25}}\,{\text{g/c}}{{\text{m}}^3}

Explanation

Solution

Use the formula for thermal expansion of a material in terms of its density. This formula gives the relation between initial density of material, changed density of material, coefficient of volumetric thermal expansion and change in temperature of material. Also, use the formula for volumetric thermal expansion coefficient in terms of linear thermal expansion coefficient.

Formula used:
The expression for thermal expansion of a material in terms of its density is given by
ρ=ρ0(1γΔT)\rho = {\rho _0}\left( {1 - \gamma \Delta T} \right) …… (1)
Here, ρ0{\rho _0} is the initial or original density of the material, ρ\rho is the changed density of material, γ\gamma is the coefficient of volumetric thermal expansion and ΔT\Delta T is the change in temperature of the material.
The volumetric thermal expansion coefficient γ\gamma is
γ=3α\gamma = 3\alpha
Here, α\alpha is the coefficient of linear thermal expansion.

Complete step by step answer:
We have given that the initial density of lead at temperature 0C0^\circ {\text{C}} is 11.34gm/cc{\text{11}}{\text{.34}}\,{\text{gm/cc}}.
Ti=0C{T_i} = 0^\circ {\text{C}}
ρ0=11.34gm/cc{\rho _0} = {\text{11}}{\text{.34}}\,{\text{gm/cc}}
We have asked to determine the changed density ρ\rho of lead at the temperature 100C100^\circ {\text{C}}.
Tf=100C{T_f} = 100^\circ {\text{C}}
The coefficient of linear thermal expansion for lead is 28×106/C28 \times {10^{ - 6}}/^\circ {\text{C}}.
α=28×106/C\alpha = 28 \times {10^{ - 6}}/^\circ {\text{C}}
We should first determine the temperature difference between two temperatures of lead.
ΔT=TfTi\Delta T = {T_f} - {T_i}

Substitute 100C100^\circ {\text{C}} for Tf{T_f} and 0C0^\circ {\text{C}} for Ti{T_i} in the above equation.
ΔT=(100C)(0C)\Delta T = \left( {100^\circ {\text{C}}} \right) - \left( {0^\circ {\text{C}}} \right)
ΔT=100C\Rightarrow \Delta T = 100^\circ {\text{C}}
Hence, the change in temperature of the lead is 100C100^\circ {\text{C}}.
Let us now determine the volumetric coefficient of thermal expansion.
Substitute 28×106/C28 \times {10^{ - 6}}/^\circ {\text{C}} for α\alpha in equation (2).
γ=3(28×106/C)\gamma = 3\left( {28 \times {{10}^{ - 6}}/^\circ {\text{C}}} \right)
γ=84×106/C\Rightarrow \gamma = 84 \times {10^{ - 6}}/^\circ {\text{C}}
Hence, the coefficient of volumetric thermal expansion is 84×106/C84 \times {10^{ - 6}}/^\circ {\text{C}}.
Let us determine the changed density of lead using equation (1).
Substitute 11.34gm/cc{\text{11}}{\text{.34}}\,{\text{gm/cc}} for ρ0{\rho _0}, 84×106/C84 \times {10^{ - 6}}/^\circ {\text{C}} for γ\gamma and 100C100^\circ {\text{C}} for ΔT\Delta T in equation (1).
ρ=(11.34gm/cc)[1(84×106/C)(100C)]\rho = \left( {{\text{11}}{\text{.34}}\,{\text{gm/cc}}} \right)\left[ {1 - \left( {84 \times {{10}^{ - 6}}/^\circ {\text{C}}} \right)\left( {100^\circ {\text{C}}} \right)} \right]
ρ=(11.34gm/cc)[184×104]\Rightarrow \rho = \left( {{\text{11}}{\text{.34}}\,{\text{gm/cc}}} \right)\left[ {1 - 84 \times {{10}^{ - 4}}} \right]
ρ=(11.34gm/cc)[10.0084]\Rightarrow \rho = \left( {{\text{11}}{\text{.34}}\,{\text{gm/cc}}} \right)\left[ {1 - 0.0084} \right]
ρ=(11.34gm/cc)(0.9916)\Rightarrow \rho = \left( {{\text{11}}{\text{.34}}\,{\text{gm/cc}}} \right)\left( {0.9916} \right)
ρ=11.24g/cm3\Rightarrow \rho = {\text{11}}{\text{.24}}\,{\text{g/c}}{{\text{m}}^3}
ρ11.25g/cm3\Rightarrow \rho \approx {\text{11}}{\text{.25}}\,{\text{g/c}}{{\text{m}}^3}

Therefore, the density of lead is 11.25g/cm3{\text{11}}{\text{.25}}\,{\text{g/c}}{{\text{m}}^3}.

So, the correct answer is “Option D”.

Note:
The students may forget to multiply the coefficient of linear thermal expansion by 3. As the density of the lead has changed, the volume of the lead also changes. Hence, the term coefficient of linear thermal expansion is replaced by volumetric thermal expansion coefficient which is thrice the linear thermal expansion coefficient.