To determine the minimum temperature to which the copper ring must be heated in order to fit over the steel rod, we can use the concept of thermal expansion.
The area \( A \) of a material expands with temperature according to the equation:
\[ A_t = A_0 (1 + \alpha \Delta T) \]
where:
- \( A_t \) is the area after temperature change,
- \( A_0 \) is the original area,
- \( \alpha \) is the coefficient of linear thermal expansion for the material,
- \( \Delta T \) is the change in temperature.
For a circular copper ring, the coefficient of linear thermal expansion \( \alpha \) is approximately \( 16 \times 10^{-6} , \text{°C}^{-1} \).
Given:
- The initial area of the hole in the copper ring, \( A_0 = 9.980 , \text{cm}^2 \).
- The area of the steel rod, \( A_s = 10.0 , \text{cm}^2 \).
To find the minimum temperature \( T \) that the copper must reach, we want the expanded area \( A_t \) of the hole in the copper ring to be at least equal to the cross-sectional area of the rod (\( 10.0 , \text{cm}^2 \)). Therefore, we set up the equation:
\[ A_0 (1 + \alpha \Delta T) = A_s \]
Substituting in the given values:
\[ 9.980 , \text{cm}^2 \cdot (1 + 16 \times 10^{-6} \Delta T) = 10.0 , \text{cm}^2 \]
Expanding the equation gives:
\[ 9.980 + (9.980 \cdot 16 \times 10^{-6} \Delta T) = 10.0 \]
Now, isolating \( \Delta T \):
\[ (9.980 \cdot 16 \times 10^{-6} \Delta T) = 10.0 - 9.980 \] \[ 9.980 \cdot 16 \times 10^{-6} \Delta T = 0.020 \] \[ \Delta T = \frac{0.020}{9.980 \cdot 16 \times 10^{-6}} \]
Calculating \( \Delta T \):
\[ \Delta T = \frac{0.020}{0.00015968} \approx 125.2 , \text{°C} \]
Now, to find the final temperature \( T \):
\[ T = 20 , \text{°C} + \Delta T \] \[ T \approx 20 , \text{°C} + 125.2 , \text{°C} \approx 145.2 , \text{°C} \]
Therefore, the minimum temperature to which the copper ring must be heated to fit over the steel rod is approximately:
\[ \boxed{145.2 , \text{°C}} \]