Determining Temperature to Expand Steel Plate Hole for Cylinder Fit

a cylinder of diameter 1 cm at 30 degrees Celsius is to slid into a hole on a steel plate. the whole has a diameter of .9997 cm at 30 degrees Celsius. to what temperature must the plate be heated Coeficient of thermal expansion C=12x10-6/deg C

To determine the temperature to which the steel plate must be heated so that the hole expands enough to accommodate the cylinder, we can use the formula for linear thermal expansion:

[ \Delta L = L_0 \alpha \Delta T ]

where:

• (\Delta L) is the change in length (or diameter, in this case),
• (L_0) is the original length (or diameter),
• (\alpha) is the coefficient of thermal expansion,
• (\Delta T) is the change in temperature.

Given:

• The original diameter of the hole, (L_0 = 0.9997 , \text{cm}),
• The desired diameter of the hole, (L = 1.0000 , \text{cm}),
• The coefficient of thermal expansion, (\alpha = 12 \times 10^{-6} , \text{per degree Celsius}),
• The initial temperature, (T_0 = 30 , \text{°C}).

First, calculate the change in diameter needed:

[ \Delta L = L - L_0 = 1.0000 , \text{cm} - 0.9997 , \text{cm} = 0.0003 , \text{cm} ]

Next, use the linear thermal expansion formula to find the change in temperature:

[ \Delta L = L_0 \alpha \Delta T ]

Rearrange to solve for (\Delta T):

[ \Delta T = \frac{\Delta L}{L_0 \alpha} ]

Substitute the known values:

[ \Delta T = \frac{0.0003 , \text{cm}}{0.9997 , \text{cm} \times 12 \times 10^{-6} , \text{per degree Celsius}} ]

[ \Delta T = \frac{0.0003}{0.9997 \times 12 \times 10^{-6}} ]

[ \Delta T \approx \frac{0.0003}{11.9964 \times 10^{-6}} ]

[ \Delta T \approx 25.01 , \text{°C} ]

Finally, add the change in temperature to the initial temperature to find the final temperature:

[ T_f = T_0 + \Delta T ]

[ T_f = 30 , \text{°C} + 25.01 , \text{°C} ]

[ T_f \approx 55.01 , \text{°C} ]

Therefore, the steel plate must be heated to approximately 55.01 degrees Celsius for the hole to expand enough to accommodate the cylinder.