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Zorluk: KolayThermal Expansion of Solids (Linear, Area, and Volume Expansivity)

A metallic rod of initial length 2.0 m2.0\text{ m} experiences a temperature increase of 50 K50\text{ K}. If the linear expansivity of the metal is 1.5×10−5 K−11.5 \times 10^{-5}\text{ K}^{-1}, what is the expansion in length of the rod in millimetres?

Cevap: 1.5 mm

Cevap

The expansion in length of the rod is 1.5 mm.
The expansion in length is calculated using ΔL=L0αΔT\Delta L = L_0 \alpha \Delta T. Substituting the values L0=2.0 mL_0 = 2.0\text{ m}, α=1.5×10−5 K−1\alpha = 1.5 \times 10^{-5}\text{ K}^{-1}, and ΔT=50 K\Delta T = 50\text{ K} gives ΔL=1.5×10−3 m\Delta L = 1.5 \times 10^{-3}\text{ m}, which equals 1.5 mm1.5\text{ mm}.

Adım Adım Çözüm

1
Identify known variables from the problem statement.
L0=2.0 mL_0 = 2.0\text{ m}, ΔT=50 K\Delta T = 50\text{ K}, and α=1.5×10−5 K−1\alpha = 1.5 \times 10^{-5}\text{ K}^{-1}.
Listing given physical quantities clarifies which thermal expansion formula to apply.
2
Calculate the change in length in metres using ΔL=L0αΔT\Delta L = L_0 \alpha \Delta T.
ΔL=2.0 m×(1.5×10−5 K−1)×50 K=0.0015 m\Delta L = 2.0\text{ m} \times (1.5 \times 10^{-5}\text{ K}^{-1}) \times 50\text{ K} = 0.0015\text{ m}.
Thermal expansion in one dimension is directly proportional to initial length, linear expansivity, and temperature change.
3
Convert the calculated expansion from metres to millimetres.
0.0015 m×1000 mm/m=1.5 mm0.0015\text{ m} \times 1000\text{ mm/m} = 1.5\text{ mm}.
The question explicitly requests the value in millimetres.

Anahtar Kavram

Linear thermal expansivity defines the fractional change in length per degree temperature change.
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Konular

Thermal Expansion of Solids (Linear, Area, and Volume Expansivity)

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