Verified answer

A messy algebraic manipulation question. You have to solve for α.

So let 65.5Ω at 54°C be R_T₁ at T₁ and 39.7Ω at 26°C be R_T₂ at T₂. It does not matter which one is which, I did this to eliminate negatives when I do calculation.

So R_T = (1 + α T) R_0. But we can do a ratio R_T₂ / R_T₁, which eliminates R_0.

R_T₂ / R_T₁ = (1 + α T₂) / (1 + α T₁)

R_T₂ (1 + α T₁) = R_T₁ (1 + α T₂)

R_T₂ + α T₁ R_T₂ = R_T₁ + α T₂ R_T₁

α T₁ R_T₂ - α T₂ R_T₁ = R_T₁ - R_T₂

α (T₁ R_T₂ - T₂ R_T₁) = R_T₁ - R_T₂

α = (R_T₁ - R_T₂) / (T₁ R_T₂ - T₂ R_T₁)

α = (65.5Ω - 39.7Ω) / (54°C * 39.7Ω - 26°C * 65.5Ω) = 0.0585

Which seems high. Cu α = 0.00427