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Question: A metal sheet having size of \(0.6 \times 0.5 m\) is heated from \[293{\text{ }}K\] to \(520^\circ C...

A metal sheet having size of 0.6×0.5m0.6 \times 0.5 m is heated from 293 K293{\text{ }}K to 520∘C520^\circ C. The final area of the hot sheet is \left\\{ {{\alpha_{metal}} = 2 \times {{10}^{ - 5}}/^\circ C} \right\\}
A. 0.306m20.306{m^2}
B. 0.0306m20.0306{m^2}
C. 3.06m23.06{m^2}
D. 1.02m21.02{m^2}

Explanation

Solution

As in the given question, initial and final temperature is given. We need to calculate the area of the hot sheet. When the temperature increases then the area will also increase.
By using the given formula, we can calculate the area of the metal sheet.
A2=A(1+β×(t2−t1)){A_2} = A(1 + \beta \times ({t_2} - {t_1}))
Here A2A_2 is the final area and (t2−t1t_2-t_1) is the change in temperature, β\beta is the thermal expansion coefficient.

Complete step by step solution:
Given: Initial temperature t1=293K{t_1} = 293K
Final temperature t2={t_2} = 520∘C520^\circ C = 793 K793{\text{ }}K
Length of metal sheet = 0.6 m0.6{\text{ }}m
Breadth of metal sheet = 0.5 m0.5{\text{ }}m
αmetal=2×10−5/∘C{\alpha _{metal}} = 2 \times {10^{ - 5}}/^\circ C
Metals are good conductors of heat and electricity. When heat is given to any metal, the particles of metals start vibrating to conduct electricity from hot area to cold area. These atomic vibrations cause expansion in the area of metal or thermal expansion. Since we know,
β=2×α\beta = 2 \times \alpha
We can calculate β\beta
∴β=2×2×10−5\therefore \beta = 2 \times 2 \times {10^{ - 5}}
∴β=4×10−5\therefore \beta = 4 \times {10^{ - 5}} ...............………………………… (I)
Now, the difference between the temperatures is given by subtracting initial temperature from final temperature.
∴t2−t1=793−293=500\therefore {t_2} - {t_1} = 793 - 293 = 500 …………………………………..(II)
Now area of metal sheet initially (A) = Length of metal sheet × Breadth of metal sheetLength{\text{ }}of{\text{ }}metal{\text{ }}sheet{\text{ }} \times {\text{ }}Breadth{\text{ }}of{\text{ }}metal{\text{ }}sheet
area of metal sheet initially (A) = 0.6 × 0.50.6{\text{ }} \times {\text{ }}0.5
area of metal sheet initially(A) = 0.3m20.3{m^2} …………….(III)
When the temperature is increased the length, breadth, area and volume of a metal, is automatically increased. This process is called thermal expansion.

Area of hot sheet when temperature is increased =A2=A(1+β×(t2−t1)){A_2} = A(1 + \beta \times ({t_2} - {t_1}))……………….(IV)
Put values of equation (I), (II) and (III) in equation (IV):
⇒A2=0.3×(1+(4×10−5)×500)\Rightarrow {A_2} = 0.3 \times (1 + (4 \times {10^{ - 5}}) \times 500)
⇒A2=0.3×1.02\Rightarrow {A_2} = 0.3 \times 1.02
⇒A2=0.306m2\Rightarrow {A_2} = 0.306{m^2}
Hence when a metal sheet of area 0.3m20.3{m^2} is heated from 293 K to 520∘C520^\circ C, the area of metal sheet will be: A2=0.306m2{A_2} = 0.306{m^2}

Therefore, option (A) is the correct answer.

Note: When the temperature is increased the length, breadth, area and volume of a metal, is automatically increased. This process is called thermal expansion. Due to the thermal expansion of a ferromagnetic metal, it reduces its magnetization, and leads to its complete loss of its magnetism.