In a certain bimetallic strip, the brass strip is 0.100% longer than the steel strip at a temperature of 295°C. At what temperature do the two strips have the same length? Coefficients of linear expansion for steel α = 12.0 × 10−6 K−1 and for brass α = 19.0 × 10−6 K−1
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To find the temperature at which the two strips have the same length, we can use the concept of linear expansion.
The linear expansion of a material can be described by the equation: ΔL = L₀αΔT, where ΔL is the change in length, L₀ is the initial length, α is the coefficient of linear expansion, and ΔT is the change in temperature.
In this case, we are given that the brass strip is 0.100% longer than the steel strip at a temperature of 295°C. This can be expressed as ΔL = 0.001L₀, where L₀ is the initial length of the steel strip.
Now, let's assume that at a certain temperature T, the two strips have the same length. We can set up the equation:
0.001L₀ = L₀(12.0 × 10^-6)(T - 295) (1 - the coefficient for the brass)
Simplifying the equation, we get:
0.001 = 12.0 × 10^-6(T - 295)
0.001/(12.0 × 10^-6) = T - 295
T = (0.001/(12.0 × 10^-6)) + 295
Now, we can calculate the value of T using the given coefficients of linear expansion for steel and brass:
T = (0.001/(12.0 × 10^-6)) + 295
T = 83.33 + 295
T ≈ 378.33°C
Therefore, the temperature at which the two strips have the same length is approximately 378.33°C.
To find the temperature at which the two strips have the same length, we can use the principle of thermal expansion.
Let's assume the initial length of the steel strip is L. Then, the initial length of the brass strip would be L + 0.100% of L, or L + 0.001L.
Let's also assume the temperature at which the two strips have the same length is T.
Using the coefficients of linear expansion, we can calculate the change in length for each strip as:
For steel: ΔL_s = α_s * L * ΔT = (12.0 × 10^(-6) K^(-1)) * L * (T - 295)
For brass: ΔL_b = α_b * (L + 0.001L) * ΔT = (19.0 × 10^(-6) K^(-1)) * (L + 0.001L) * (T - 295)
Since the two strips have the same length at this temperature, we can set ΔL_s = -ΔL_b:
(12.0 × 10^(-6) K^(-1)) * L * (T - 295) = - (19.0 × 10^(-6) K^(-1)) * (L + 0.001L) * (T - 295)
Now, let's solve for T:
12.0 × 10^(-6) * (T - 295) = -19.0 × 10^(-6) * (1.001) * (T - 295)
12.0 × 10^(-6)T - 12.0 × 10^(-6) * 295 = -19.0 × 10^(-6) * 1.001T + 19.0 × 10^(-6) * 1.001 * 295
12.0 × 10^(-6)T - 12.0 × 10^(-6) * 295 = -19.0 × 10^(-6)T + 19.0 × 10^(-6) * 295
12.0 × 10^(-6)T + 19.0 × 10^(-6)T = 12.0 × 10^(-6) * 295 + 19.0 × 10^(-6) * 295
31.0 × 10^(-6)T = 12.0 × 10^(-6) * 295 + 19.0 × 10^(-6) * 295
31.0 × 10^(-6)T = (12.0 + 19.0) × 10^(-6) * 295
31.0T = 31.0 * 295
T = (31.0 * 295) / 31.0
T = 295
Therefore, the two strips have the same length at a temperature of 295°C.