A steel rod 1m at 20 degrees C — find length at 100 degrees C

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Tags Thermal Expansion

Question

A steel rod has a length of 1 m at 20°C. Find its length at 100°C. (Coefficient of linear expansion of steel, α=12×10−6\alpha = 12 \times 10^{-6} °C⁻¹)

Solution — Step by Step

We have initial length L0=1L_0 = 1 m, initial temperature T1=20°CT_1 = 20°C, final temperature T2=100°CT_2 = 100°C, and α=12×10−6\alpha = 12 \times 10^{-6} °C⁻¹.

The temperature change is ΔT=T2−T1=100−20=80°C\Delta T = T_2 - T_1 = 100 - 20 = 80°C.

The increase in length due to thermal expansion is given by:

ΔL=L0⋅α⋅ΔT\Delta L = L_0 \cdot \alpha \cdot \Delta T

This formula says: the rod expands by a fraction α\alpha of its original length for every degree rise in temperature. The key insight here is that α\alpha is a material property — steel expands at a fixed rate per degree.

Substituting values:

ΔL=1×12×10−6×80\Delta L = 1 \times 12 \times 10^{-6} \times 80 ΔL=960×10−6 m=9.6×10−4 m\Delta L = 960 \times 10^{-6} \text{ m} = 9.6 \times 10^{-4} \text{ m}

The final length is:

L=L0+ΔL=1+9.6×10−4L = L_0 + \Delta L = 1 + 9.6 \times 10^{-4} L=1.00096 m\boxed{L = 1.00096 \text{ m}}

This can also be written as L=L0(1+αΔT)=1×(1+12×10−6×80)=1.00096L = L_0(1 + \alpha \Delta T) = 1 \times (1 + 12 \times 10^{-6} \times 80) = 1.00096 m.

Why This Works

Thermal expansion happens because atoms in a solid vibrate more vigorously at higher temperatures. They push each other slightly farther apart — the bond length increases on average. The coefficient α\alpha captures how sensitive this material is to temperature changes.

Steel has a relatively small α\alpha (compared to, say, aluminium at 24×10−624 \times 10^{-6} °C⁻¹), which is why steel bridges can withstand large temperature swings without cracking — provided expansion joints are included to allow free movement.

The formula L=L0(1+αΔT)L = L_0(1 + \alpha \Delta T) is a linear approximation valid for small ΔT\Delta T. For very large temperature changes, higher-order terms matter, but for most practical problems in Class 11 and JEE, this formula is exact enough.

Alternative Method

We can directly use the combined formula without separating ΔL\Delta L:

L=L0(1+αΔT)L = L_0(1 + \alpha \Delta T) L=1×(1+12×10−6×80)=1×1.00096=1.00096 mL = 1 \times (1 + 12 \times 10^{-6} \times 80) = 1 \times 1.00096 = 1.00096 \text{ m}

This is slightly faster and less prone to arithmetic errors.

Common Mistake

Many students substitute the temperatures directly as T1T_1 and T2T_2 into some modified formula, or forget that ΔT=T2−T1\Delta T = T_2 - T_1 and not T2T_2 alone. Always calculate ΔT\Delta T first. Also, watch the units: α\alpha is in °C⁻¹, so ΔT\Delta T must be in °C (not Kelvin — though the numerical difference is the same, it helps to be consistent).

For JEE, remember that the same numerical value of ΔT\Delta T holds whether you use Celsius or Kelvin (since they differ only by an additive constant, not a scaling factor). So ΔTCelsius=ΔTKelvin\Delta T_{Celsius} = \Delta T_{Kelvin} always.

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