original length=2.0m, linear expansivity=25(10'-6 k'-1),rise in temp=?,increase in length=3.0 (cal the rise in temp)
changelength=origLength(alpha)(deltatemp)
you are asking for an original length of 2m, and increase of 3m to 5m? I dont believe that data.
To calculate the rise in temperature, we can use the formula for linear expansion:
ΔL = α * L * ΔT
Where:
ΔL is the increase in length
α is the linear expansivity
L is the original length
ΔT is the rise in temperature
We are given:
ΔL = 3.0m
L = 2.0m
α = 25(10^(-6) K^(-1))
Substituting these values into the formula, we get:
3.0m = (25(10^(-6) K^(-1))) * (2.0m) * ΔT
Simplifying the equation, we have:
3.0m = (50(10^(-6) K^(-1))) * ΔT
Now, let's solve for ΔT:
ΔT = 3.0m / (50(10^(-6) K^(-1)))
ΔT = 3.0m / (5 * 10^(-4) K^(-1))
ΔT = (3.0m * 10^4 K) / 5
ΔT = 6.0 * 10^4 K
Therefore, the rise in temperature is 6.0 * 10^4 K.
To find the rise in temperature when the increase in length is given, we can use the formula for linear expansion:
ΔL = αLΔT
Where:
ΔL is the change in length,
α is the linear expansivity,
L is the original length, and
ΔT is the change in temperature.
We are given:
Original length (L) = 2.0 m
Linear expansivity (α) = 25 × 10^-6 k^-1
Increase in length (ΔL) = 3.0 m
We can rearrange the formula to solve for ΔT:
ΔT = ΔL / (αL)
Substituting the given values:
ΔT = 3.0 m / (25 × 10^-6 k^-1 × 2.0 m)
Now let's calculate the rise in temperature:
ΔT = 3.0 m / (50 × 10^-6)
Simplifying further:
ΔT = 3.0 m / (5 × 10^-5)
Now, dividing the numerator and denominator by 0.01 to convert from m to cm:
ΔT = 300 cm / 5
Finally:
ΔT = 60 °C
Therefore, the rise in temperature is 60 °C.