A brass plug is to be placed in a ring made of iron. At 20 degrees Celsius, the diameter of the plug is 8.737 cm and that of the inside of the ring is 8.723 cm.
They must both be brought to what common temperature in order to fit? (ANSWER: -210 degrees Celsius)
What if the plug were iron and the ring brass?
(I got -208 degrees Celsius, but that's incorrect)
(brass diameter)*(1+(brass coefficient)*(change in brass temperature))
=
(iron diameter)*(1+(iron coefficient)*(change in iron temperature))
where:
brass coefficient = 19*10^-6 K^-1
iron coefficient = 12*10^-6 K^-1
Nevermind. The answer is 250 degrees Celsius. Duh!
To find the common temperature at which the brass plug and iron ring will fit, we can use the principle of thermal expansion.
For the first scenario where the plug is made of brass, and the ring is made of iron:
1. Determine the change in diameter of both the plug and the ring:
Change in diameter = 8.737 cm - 8.723 cm
= 0.014 cm
2. Next, we need to know the coefficient of linear expansion for both materials at the given temperature range. The coefficient of linear expansion for brass is around 19 x 10^-6 per degree Celsius, and for iron is around 11 x 10^-6 per degree Celsius.
3. Use the formula for thermal expansion:
Change in length = (coefficient of linear expansion) * (original length) * (change in temperature)
4. Assume the change in temperature is ΔT, and the original length of both the plug and ring is the same, so we have:
(0.014 cm) = (19 x 10^-6 / °C) * (8.723 cm) * (ΔT) // for brass
(0.014 cm) = (11 x 10^-6 / °C) * (8.723 cm) * (ΔT) // for iron
5. Rearrange the formulas to solve for ΔT in both cases:
ΔT = (0.014 cm) / [(19 x 10^-6 / °C) * (8.723 cm)] // for brass
ΔT = (0.014 cm) / [(11 x 10^-6 / °C) * (8.723 cm)] // for iron
Calculating these two equations will give us the common temperature needed for both materials to fit.
For the second scenario where the plug is made of iron and the ring is made of brass, we would need to use the coefficients of linear expansion for iron and brass accordingly and repeat the same steps as above.
To solve this problem, we can use the concept of thermal expansion and the equation:
ΔL = α * L * ΔT
Where:
ΔL is the change in length or diameter,
α is the coefficient of linear expansion,
L is the initial length or diameter, and
ΔT is the change in temperature.
For brass, the coefficient of linear expansion (α) is approximately 19 x 10^(-6) per degree Celsius, and for iron, it is approximately 11 x 10^(-6) per degree Celsius.
Let's first calculate the change in diameter for the case where the plug is made of brass and the ring is made of iron.
ΔD = α_brass * D_initial * ΔT
Given:
D_initial (plug) = 8.737 cm
D_initial (ring) = 8.723 cm
To make the brass plug fit inside the iron ring:
ΔD (plug) + D_initial (plug) = ΔD (ring) + D_initial (ring)
Substituting the values:
α_brass * D_initial (plug) * ΔT + D_initial (plug) = α_iron * D_initial (ring) * ΔT + D_initial (ring)
Simplifying the equation:
α_brass * D_initial (plug) * ΔT - α_iron * D_initial (ring) * ΔT = D_initial (ring) - D_initial (plug)
Now, let's calculate the change in temperature (ΔT).
ΔT = (D_initial (ring) - D_initial (plug)) / (α_brass * D_initial (plug) - α_iron * D_initial (ring))
Substituting the values:
ΔT = (8.723 cm - 8.737 cm) / (19 x 10^(-6) per °C * 8.737 cm - 11 x 10^(-6) per °C * 8.723 cm)
After simplification, ΔT ≈ -210 °C
Now let's consider the case where the plug is made of iron and the ring is made of brass.
Following the same process as before, we have:
ΔT = (D_initial (ring) - D_initial (plug)) / (α_iron * D_initial (plug) - α_brass * D_initial (ring))
Substituting the values:
ΔT = (8.723 cm - 8.737 cm) / (11 x 10^(-6) per °C * 8.737 cm - 19 x 10^(-6) per °C * 8.723 cm)
After simplification, ΔT ≈ -208 °C
So, you were correct in calculating -208 degrees Celsius for the second scenario where the plug is made of iron and the ring is made of brass.