Examkin

Question

Difficulty: EasyThermal 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?

Answer: 1.5 mm

Answer

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}.

Step-by-Step Solution

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.

Key Concept

Linear thermal expansivity defines the fractional change in length per degree temperature change.
Rate this question

Topics

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

More questions from this topic