Lsi = 1000 meters
Lbi = 1001 meters
delta Ls = 1000 * 12.0 * 10^-6 * dT
delta Lb = 1001 * 19.0 * 10^-6 * dT
but
delta Ls = 1 + delta Lb
delta Ls = 1000 * 12.0 * 10^-6 * dT = 1 + 1001 * 19.0 * 10^-6 * dT
dT is going to be negative of course
Lbi = 1001 meters
delta Ls = 1000 * 12.0 * 10^-6 * dT
delta Lb = 1001 * 19.0 * 10^-6 * dT
but
delta Ls = 1 + delta Lb
delta Ls = 1000 * 12.0 * 10^-6 * dT = 1 + 1001 * 19.0 * 10^-6 * dT
dT is going to be negative of course
(12000 - 19019) dT = 10^6
- 7019 dT = 1,000,000
dT = - 142
so 295 - 142 = 153 deg
1. First, let's assume that the initial length of the steel strip is "L." Since the brass strip is 0.100% longer, its initial length would be L + 0.001L = 1.001L.
2. Next, we need to calculate the change in length for both materials. The change in length (∆L) for any material can be given by the formula: ∆L = α * L * ∆T, where α is the coefficient of linear expansion, L is the initial length, and ∆T is the change in temperature.
3. Let's calculate the change in length for both the steel and brass strips:
For steel: ∆L_steel = α_steel * L * ∆T
For brass: ∆L_brass = α_brass * (1.001L) * ∆T (since the initial length of brass is 1.001L)
4. Now, we can set the change in length of the steel strip equal to the change in length of the brass strip and solve for the temperature (∆T) at which they are equal. This can be expressed as:
α_steel * L * ∆T = α_brass * (1.001L) * ∆T
5. Simplifying the equation, we get:
α_steel * L = α_brass * (1.001L)
6. Now, let's substitute the given values for α_steel and α_brass:
(12.0 × 10^(-6) K^(-1)) * L = (19.0 × 10^(-6) K^(-1)) * (1.001L)
7. Cancel out the "L" term on both sides of the equation:
12.0 × 10^(-6) = 19.0 × 10^(-6) * 1.001
8. Solve for the temperature (∆T):
∆T = (19.0 × 10^(-6) * 1.001) / 12.0 × 10^(-6)
9. Calculate the value of ∆T:
∆T = 1.59018
10. Therefore, the two strips will have the same length when the temperature (∆T) is approximately 1.59018 degrees Celsius.
Remember to use the appropriate units when substituting values into the equation and perform the calculations accurately to get an accurate answer.