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Question: A thin Brass sheet at \({10^ \circ }C\)and thin steel sheet at \({20^ \circ }C\) have the same surfa...

A thin Brass sheet at 10∘C{10^ \circ }Cand thin steel sheet at 20∘C{20^ \circ }C have the same surface area. Find the common temperature at which both would have the same area:
(linear expansion coefficient for brass and steel are 19×10−6/∘C19 \times {10^{ - 6}}{/^ \circ }C and 11×10−6/∘C11 \times {10^{ - 6}}{/^ \circ }C respectively.)

(a) −3.75∘C - {3.75^ \circ }C
(b) −2.75∘C - {2.75^ \circ }C
(c) 2.75∘C{2.75^ \circ }C
(d) 3.75∘C{3.75^ \circ }C

Explanation

Solution

Derive the areal expansion coefficient from linear expansion coefficient. Then, use the thermal expansion relation to get expansion of the sheet for both the bodies.

Formula Used:
1.expansion produced due to given expansion coefficients is given by
A(T)=A0(1+β(T−T0))A(T) = {A_0}(1 + \beta (T - {T_0})) …… (1)
Where,
A(T) =area at any temperature T
T= Temperature (It is variable)
T0{T_0}= initial temperature of the respective sheet
β\beta = Coefficient of areal expansion for the sheet.

2. Relation between Coefficient of area expansion and coefficient of linear expansion is given by: β=2α\beta = 2\alpha …… (2)
Where,
β\beta = Coefficient of areal expansion for the sheet.
α\alpha = Coefficient of linear expansion for the sheet along each dimension.

Complete step by step answer:
Given,
Initial area of brass sheet: A0{A_0}
Coefficient of linear thermal expansion for the brass sheet: αb=19×10−6/∘C{\alpha _b} = 19 \times {10^{ - 6}}{/^ \circ }C .
initial temperature of the brass sheet: T0{T_0}=10∘C{10^ \circ }C

Initial area of steel sheet: A0{A_0} (initially they have same area)
Coefficient of linear thermal expansion for the steel sheet: αs=11×10−6/∘C{\alpha _s} = 11 \times {10^{ - 6}}{/^ \circ }C
initial temperature of the steel sheet: T0{T_0}=20∘C{20^ \circ }C
Let the final new temperature of the two bodies be Tf{T_f} where both bodies have the same area.

Step 1:
Using equation (2) we can find a real expansion coefficient from a given linear expansion coefficient.
For brass sheet:
Coefficient of areal thermal expansion for the brass sheet:
βb=2×αb=2×19×10−6/∘C{\beta _b} = 2 \times {\alpha _b} = 2 \times 19 \times {10^{ - 6}}{/^ \circ }C. …… (3)
Similarly, for steel sheet:
Coefficient of areal thermal expansion for the steel sheet:
βs=2×αs=2×11×10−6/∘C{\beta _s} = 2 \times {\alpha _s} = 2 \times 11 \times {10^{ - 6}}{/^ \circ }C. …… (4)

Step 2:
Using equation (1) and (2) we get areal expansion for both metal sheets-
For brass sheet:
Putting value given above for the brass sheet in equation (1) we get-
A(T)=A0(1+β(T−T0))A(T) = {A_0}(1 + \beta (T - {T_0}))
⇒A(T)b=A0(1+2×19×10−6(T−10))\Rightarrow A{(T)_b} = {A_0}(1 + 2 \times 19 \times {10^{ - 6}}(T - 10)) ……. (5)
For steel sheet:
Putting value given above for the brass sheet in equation (1) we get-
A(T)=A0(1+β(T−T0))A(T) = {A_0}(1 + \beta (T - {T_0}))
⇒A(T)s=A0(1+2×11×10−6(T−20))\Rightarrow A{(T)_s} = {A_0}(1 + 2 \times 11 \times {10^{ - 6}}(T - 20)) ……. (6)

Step 3:
Dividing equation (5) by (6) we get,

⇒A(T)bA(T)s=A0(1+2×19×10−6(T−10))A0(1+2×11×10−6(T−20)) \Rightarrow \dfrac{{A{{(T)}_b}}}{{A{{(T)}_s}}} = \dfrac{{{A_0}(1 + 2 \times 19 \times {{10}^{ - 6}}(T - 10))}}{{{A_0}(1 + 2 \times 11 \times {{10}^{ - 6}}(T - 20))}}
⇒A(T)bA(T)s=(1+2×19×10−6(T−10))(1+2×11×10−6(T−20))\Rightarrow \dfrac{{A{{(T)}_b}}}{{A{{(T)}_s}}} = \dfrac{{(1 + 2 \times 19 \times {{10}^{ - 6}}(T - 10))}}{{(1 + 2 \times 11 \times {{10}^{ - 6}}(T - 20))}} ……. (7)

Step 4:
We know finally both have same temperature Tf{T_f} and same area
SoA(T)bA(T)s\dfrac{{A{{(T)}_b}}}{{A{{(T)}_s}}}=1 ……(8)
Putting final temperature and values from equation (8) in (7) we get-
⇒1=(1+2×19×10−6(Tf−10))(1+2×11×10−6(Tf−20))\Rightarrow 1 = \dfrac{{(1 + 2 \times 19 \times {{10}^{ - 6}}({T_f} - 10))}}{{(1 + 2 \times 11 \times {{10}^{ - 6}}({T_f} - 20))}}
⇒(1+2×11×10−6(Tf−20))=(1+2×19×10−6(Tf−10))\Rightarrow (1 + 2 \times 11 \times {10^{ - 6}}({T_f} - 20)) = (1 + 2 \times 19 \times {10^{ - 6}}({T_f} - 10))
Simplifying both sides, we get-

⇒11(Tf−20)=19(Tf−10) ⇒−220+190=(19−11)Tf ⇒−30=8Tf ⇒Tf=−308=−3.75∘C  \Rightarrow 11({T_f} - 20) = 19({T_f} - 10) \\\ \Rightarrow - 220 + 190 = (19 - 11){T_f} \\\ \Rightarrow - 30 = 8{T_f} \\\ \Rightarrow {T_f} = \dfrac{{ - 30}}{8} = - {3.75^ \circ }C \\\

Final Answer
Hence, option (a) −3.75∘C - {3.75^ \circ }C is correct.

Note: It is important to convert coefficient of linear expansion into coefficient of areal expansion, as there are three different kinds of expansion coefficient for a material i.e Linear, Areal and Volumetric.