Some textbooks use the unit \(\mathrm{K}^{-1}\) rather than \({ }^{\circ} \mathrm{C}^{-1}\) for values of the linear expansion coefficient; see Table 17.2 How will the numerical values of the coefficient differ if expressed in \(\mathrm{K}^{-1}\) ?

Short Answer

Expert verified
Answer: No, there is no difference between the linear expansion coefficient when expressed in K⁻¹ and °C⁻¹. This is because both Kelvin and Celsius scales have the same incremental difference in temperature, which influences the numerical value of the linear expansion coefficient.

Step by step solution

01

Relationship of temperature scales

Both the Celsius (C) and Kelvin (K) scales of temperature have the same incremental difference (1 degree), but their numbering systems start from different points: absolute zero (0 K) for the Kelvin scale and the freezing point of water (0 °C) for the Celsius scale. The relationship between these two scales can be expressed as: T(K) = T(°C) + 273.15
02

Analyzing the expression of linear expansion coefficient in both units

The linear expansion coefficient, α, describes the tendency of a solid to expand or contract as a function of temperature changes. Mathematically, it is given by: ΔL = L₀αΔT, where ΔL is the change in length, L₀ is the initial length, and ΔT is the change in temperature. Since the incremental difference in temperature for both Kelvin and Celsius scales is the same, we should not observe any significant difference when expressed in K⁻¹ or °C⁻¹. The linear expansion coefficient should remain the same.
03

Comparing numerical values of the same coefficient in K⁻¹ and °C⁻¹

It can be verified that the numerical values of the coefficients will not differ when expressed in K⁻¹ or °C⁻¹. This is because the main effect of the change of temperature is what influences the numerical value of the linear expansion coefficient. Since the increments are the same in both scales, the values of α in K⁻¹ and °C⁻¹ will be equal for the same temperature change.

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