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Question: The volume of a metal sphere increases by \(0.24\%\) when its temperature is raised by \({40^0}C\). ...

The volume of a metal sphere increases by 0.24%0.24\% when its temperature is raised by 400C{40^0}C. What is the coefficient of linear expansion of the metal?
A) 2×10−5∘C−12 \times {10^{ - 5}}{}^ \circ {C^{ - 1}}
B) 6×10−5∘C−16 \times {10^{ - 5}}{}^ \circ {C^{ - 1}}
C) 18×10−5∘C−118 \times {10^{ - 5}}{}^ \circ {C^{ - 1}}
D) 1.2×10−5∘C−11.2 \times {10^{ - 5}}{}^ \circ {C^{ - 1}}

Explanation

Solution

When a matter is subjected to heat or a temperature change, its atoms gain energy and move very fast, leading to thermal expansion. When we heat things, not only their length changes, but there is also a change in their area and volume. This gives rise to the need to define three different coefficients of thermal expansion, namely, linear, areal, and volumetric coefficients of thermal expansion.

Complete step by step solution:
For a small value of ΔV\Delta V:
ΔV∝ΔT\Delta V \propto \Delta T and ΔV∝V\Delta V \propto V
⇒ΔV∝VΔT\Rightarrow \Delta V \propto V\Delta T
∴ΔV=γVΔT\therefore \Delta V = \gamma V\Delta T……………….. Equation (1)
where,
γ=\gamma = coefficient of volumetric expansion
V=V = volume of body
ΔT=\Delta T = change in temperature
ΔV=\Delta V = a small change in volume [as ΔV≪V\Delta V \ll V]
Now, ΔV=VT−V0\Delta V = {V_T} - {V_0} as
V0={V_0} = the initial volume of the body and
VT={V_T} = the volume of the body after increasing temperature, i.e., at the temperature TT
⇒VT−V0=γV0ΔT\Rightarrow {V_T} - {V_0} = \gamma {V_0}\Delta T
⇒VT=V0+γV0ΔT\Rightarrow {V_T} = {V_0} + \gamma {V_0}\Delta T
Therefore, the change in the volume of a body is given by:
⇒VT=V0(1+γΔT)\Rightarrow {V_T} = {V_0}(1 + \gamma \Delta T)
Here, we have ΔVV=0.24%=0.24100\dfrac{{\Delta V}}{V} = 0.24\% = \dfrac{{0.24}}{{100}} and ΔT=40∘C\Delta T = 40{}^ \circ C
Using equation (1) and rearranging it, we get
⇒γ=ΔVVΔT\Rightarrow \gamma = \dfrac{{\Delta V}}{{V\Delta T}}
Substituting the values to find the coefficient of volumetric expansion:
⇒γ=0.24100×140\Rightarrow \gamma = \dfrac{{0.24}}{{100}} \times \dfrac{1}{{40}}
⇒γ=0.06×10−3\Rightarrow \gamma = 0.06 \times {10^{ - 3}}
⇒γ=6×10−5∘C−1\Rightarrow \gamma = 6 \times {10^{ - 5}}{}^ \circ {C^{ - 1}}
Just like we derived the coefficient of volumetric expansion, similarly according to equation (1), linear expansion of a body is expressed as:
ΔL=αLΔT\Delta L = \alpha L\Delta T……………………….Equation (2)
where α=\alpha = coefficient of linear expansion
Let us assume a cube of the side LL whose volume is L3{L^3}.
Now, ΔV=(L+ΔL)3−L3\Delta V = {(L + \Delta L)^3} - {L^3}
Using the mathematical formula, (a+b)3=a3−3a2b+3ab2−b3{(a + b)^3} = {a^3} - 3{a^2}b + 3a{b^2} - {b^3}
We get ΔV=[L3−3L(ΔL)2+3L2.ΔL−(ΔL)3]−L3\Delta V = [{L^3} - 3L{(\Delta L)^2} + 3{L^2}.\Delta L - {(\Delta L)^3}] - {L^3}
Since, ΔL≪L\Delta L \ll L,(ΔL)3≈0{(\Delta L)^3} \approx 0 and (ΔL)2≈0{(\Delta L)^2} \approx 0
⇒ΔV=L3+3L2.ΔL−L3\Rightarrow \Delta V = {L^3} + 3{L^2}.\Delta L - {L^3}
⇒ΔV=3L2.ΔL\Rightarrow \Delta V = 3{L^2}.\Delta L
Substituting the value of ΔL\Delta L from equation (2);
⇒ΔV=3L2(αLΔT)\Rightarrow \Delta V = 3{L^2}(\alpha L\Delta T)
⇒ΔV=3αL3ΔT\Rightarrow \Delta V = 3\alpha {L^3}\Delta T……….Equation (3)
Comparing equations (1) and (3), we get
⇒γ=3α\Rightarrow \gamma = 3\alpha
⇒α=γ3\Rightarrow \alpha = \dfrac{\gamma }{3}
⇒α=6×10−5∘C−13\Rightarrow \alpha = \dfrac{{6 \times {{10}^{ - 5}}{}^ \circ {C^{ - 1}}}}{3}
⇒α=2×10−5∘C−1\Rightarrow \alpha = 2 \times {10^{ - 5}}{}^ \circ {C^{ - 1}}

The correct answer is [A], 2×10−5∘C−12 \times {10^{ - 5}}{}^ \circ {C^{ - 1}}.

Note: Every material has its unique coefficient of volumetric expansion, but all three different coefficients of thermal expansion are interdependent, further indicating that these coefficients are also a characteristic property of every material (mostly metal).