Which expression correctly represents the linear expansion deltaL for a material with initial length L0, coefficient alpha, and temperature change deltaT?

Which expression correctly represents the linear expansion deltaL for a material with initial length L0, coefficient alpha, and temperature change deltaT?

• A. deltaL = alpha * deltaT
• B. deltaL = deltaT * L0
• C. deltaL = alpha * L0 * deltaT
• D. deltaL = alpha + L0 + deltaT

Answer: (C)

Linear expansion shows how much a length changes in proportion to the original length, the temperature change, and the material’s coefficient of linear expansion. The actual change in length is the original length times the fractional change, which is alpha times deltaT. So deltaL = L0 * (alpha * deltaT) = alpha * L0 * deltaT. This reflects why longer pieces change more and why bigger temperature increases or larger alpha produce a bigger deltaL. Units line up as well: alpha is 1 per degree, deltaT is degrees, and L0 is a length, giving deltaL a length unit. If deltaT is negative, the material shrinks. Expressions that omit L0 or alpha lose either the length scale or the material property, leading to an incorrect result. For example, with L0 = 2 m, alpha = 1e-5 /°C, deltaT = 50°C, deltaL = 1e-5 × 2 × 50 = 0.001 m (1 mm).

Linear expansion shows how much a length changes in proportion to the original length, the temperature change, and the material’s coefficient of linear expansion. The actual change in length is the original length times the fractional change, which is alpha times deltaT. So deltaL = L0 * (alpha * deltaT) = alpha * L0 * deltaT. This reflects why longer pieces change more and why bigger temperature increases or larger alpha produce a bigger deltaL. Units line up as well: alpha is 1 per degree, deltaT is degrees, and L0 is a length, giving deltaL a length unit. If deltaT is negative, the material shrinks. Expressions that omit L0 or alpha lose either the length scale or the material property, leading to an incorrect result. For example, with L0 = 2 m, alpha = 1e-5 /°C, deltaT = 50°C, deltaL = 1e-5 × 2 × 50 = 0.001 m (1 mm).