An iron ball of diameter 15.23cm rests on a brass ring of internal diameter 15.0cm at a temperature of 20 degrees celsius. To what temperature must both be heated for the ball to pass through the ring?

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To find the temperature at which the iron ball will pass through the brass ring, we need to consider the thermal expansion of both the iron ball and the brass ring.

The principle of thermal expansion states that when an object is heated, it expands in size. The change in size due to temperature is given by the coefficient of linear expansion (α), which is different for different materials.

To solve this problem, we need to use the equations for thermal expansion. The equation that relates the change in length (or diameter) to the change in temperature is:

ΔL = L * α * ΔT

Where:
ΔL is the change in length (or diameter),
L is the original length (or diameter),
α is the coefficient of linear expansion of the material, and
ΔT is the change in temperature.

Given:
Diameter of the iron ball (original) = 15.23 cm
Diameter of the brass ring (original) = 15.0 cm
Temperature (initial) = 20°C

Since we want the iron ball to pass through the brass ring, we need to find the temperature at which the diameter of the iron ball will be equal to the internal diameter of the brass ring.

Let's assume the final temperature we want to find is T.

Using the equation above, we can set up the following equation:

Diameter_of_iron_ball + ΔDiameter_of_iron_ball = Diameter_of_brass_ring

(15.23 cm) + [15.23 cm * α_iron * (T - 20°C)] = 15.0 cm

Now, we need to know the coefficient of linear expansion for both iron and brass. The coefficient of linear expansion for iron is approximately α_iron = 12 x 10^-6 K^-1 and the coefficient of linear expansion for brass is approximately α_brass = 18 x 10^-6 K^-1.

Substituting the values:

15.23 cm + [15.23 cm * 12 x 10^-6 K^-1 * (T - 20°C)] = 15.0 cm

Now, we can solve for T:

[15.23 cm * 12 x 10^-6 K^-1 * (T - 20°C)] = 15.0 cm - 15.23 cm

[15.23 cm * 12 x 10^-6 K^-1 * (T - 20°C)] = -0.23 cm

(T - 20°C) = (-0.23 cm) / [15.23 cm * 12 x 10^-6 K^-1]

T - 20°C = -1255.2 K

T = -1255.2 K + 20°C

T ≈ -1235.2°C

Hence, to pass through the ring, both the iron ball and the brass ring need to be heated to a temperature of approximately -1235.2°C.