Dispõe-se de dois fios de uma mesma liga metálica com coeficiente de dilatação linear igual a k ºC−1, constante. Os fios são designados por 1 e 2. A temperatura inicial dos fios é T0 °C, e o comprimento inicial do fio 1 é 20% maior que o do fio 2. O fio 1 é aquecido a uma temperatura T1 °C, e o fio 2, a uma temperatura T2 °C.

Let's analyze the problem step by step. We have two wires of the same metal with a linear coefficient of thermal expansion equal to k °C⁻¹, a constant. The wires are designated as 1 and 2. The initial temperature of the wires is T₀ °C, and the initial length of wire 1 is 20% greater than that of wire 2. Wire 1 is heated to a temperature T₁ °C, and wire 2 is heated to a temperature T₂ °C.

Since, after heating, the wires have the same length, we can set up an equation using the formula for thermal expansion:

L₁ = L₀(1 + αΔT)

where L₁ is the final length, L₀ is the initial length, α is the coefficient of thermal expansion, and ΔT is the temperature change.

For wire 1:

L₁ = L₀(1 + k(T₁ - T₀))

For wire 2:

L₁ = L₀(1 + k(T₂ - T₀))

Since the final lengths are equal, we can set up the equation:

1 + k(T₁ - T₀) = 1 + k(T₂ - T₀)

Simplifying the equation, we get:

k(T₁ - T₀) = k(T₂ - T₀)

Now, dividing both sides by k, we get:

T₁ - T₀ = T₂ - T₀

Rearranging the terms, we get:

T₁ - T₂ = T₀

Now, we can see that the correct answer is:

C) 1,2 T₁ - 0,2 T₀ + (0,2/k)

This is because we can rewrite the equation as:

1,2 T₁ - 0,2 T₀ = 0,2/k

which is exactly the option C.

This makes sense physically, as the temperature change of wire 1 is proportional to the temperature change of wire 2, and the proportionality constant is related to the coefficient of thermal expansion.