A rod made from a particular alloy is heated from 24.5°C to the boiling point of water. Its length increases by 8.15 x 10-4m. The rod is then cooled from 24.5°C to the freezing point of water. By how much does the rod shrink?
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To find out how much the rod shrinks when cooled, we first need to calculate how much it expands when heated.
The coefficient of linear expansion (α) is a property of the material, which determines how much it expands or contracts for a given change in temperature. We can use this coefficient to calculate the thermal expansion of the rod.
Let's assume the initial length of the rod is L.
The formula to calculate the thermal expansion ΔL is:
ΔL = α * L * ΔT
Where:
- ΔL is the change in length
- α is the coefficient of linear expansion
- L is the original length of the rod
- ΔT is the change in temperature
In the first scenario, the rod is heated from 24.5°C to the boiling point of water, which is 100°C. The change in temperature (ΔT) would be:
ΔT = 100°C - 24.5°C = 75.5°C
Now, let's determine the amount of expansion for the rod using the given data.
ΔL = α * L * ΔT
ΔL = (8.15 x 10^(-4)m) * L * 75.5°C
Given that ΔL = 8.15 x 10^(-4)m, we can rearrange the equation to solve for L:
L = ΔL / (α * ΔT)
Substituting the values:
L = (8.15 x 10^(-4)m) / ((8.15 x 10^(-4)m * 75.5°C)
L = 1 / 75.5 = 0.0132m
So, the initial length (L) of the rod is 0.0132m.
Now, let's move on to the second scenario where the rod is cooled from 24.5°C to the freezing point of water, which is 0°C.
ΔT = 0°C - 24.5°C = -24.5°C
Using the same formula as before, we can calculate the amount of contraction (ΔL) for the rod:
ΔL = α * L * ΔT
ΔL = (8.15 x 10^(-4)m) * 0.0132m * (-24.5°C)
ΔL = -8.15 x 10^(-4)m * 0.0132m * 24.5°C
ΔL = -0.0259 x 10^(-4)m
Therefore, the rod shrinks by approximately 2.59 x 10^(-5)m when cooled from 24.5°C to the freezing point of water.