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### 5. Why do we use mercury in a glass thermometer while the glass undergoes thermal expansion with it?
We use mercury in a glass thermometer for several reasons:
1. Wide Temperature Range: Mercury can remain in liquid form over a wide range of temperatures, from -39°C to 356°C. This makes it ideal for measuring a vast range of temperatures.
2. High Thermal Conductivity: Mercury conducts heat well, meaning it can quickly respond to changes in temperature.
3. Minimal Adhesion: Mercury does not wet glass; it forms a cohesive droplet within the glass tube. This prevents it from sticking to the walls, ensuring consistent and accurate readings.
4. Thermal Expansion: Though glass also undergoes thermal expansion, the expansion of mercury is significantly more pronounced. This means that even though glass expands, the volumetric expansion of mercury is the dominant factor influencing the reading.
5. Visibility: Mercury has a shiny, metallic appearance, making it easily visible within the glass tube.
### 6. When a metal ring and a metal sphere with equal radius are both at room temperature, the sphere does not pass through the ring. After the ring is heated, the sphere can pass through the ring. Why?
When the metal ring is heated, it undergoes thermal expansion. This means its dimensions increase, including its inner diameter. As the ring expands, the gap in the middle becomes larger, allowing the sphere with the same radius to pass through. The increase in the ring’s inner diameter due to heating compensates for the sphere’s radius that initially prevented it from passing through.
### 1. The diameter of a piston is 10 cm. To what temperature should a steel ring of diameter 9.9 cm be heated so that the piston can pass through the ring?
To solve this problem, we must consider the thermal expansion formula for solids:
[tex]\[ \Delta L = L_0 \alpha \Delta T \][/tex]
- [tex]\( \Delta L \)[/tex] is the change in length (diameter in this case).
- [tex]\( L_0 \)[/tex] is the original length (9.9 cm).
- [tex]\( \alpha \)[/tex] is the coefficient of linear expansion for steel (approx. [tex]\( 12 \times 10^{-6} / \text{°C} \)[/tex]).
- [tex]\( \Delta T \)[/tex] is the change in temperature.
For the diameter of the ring to increase from 9.9 cm to 10 cm:
[tex]\[ 10 - 9.9 = 9.9 \times 12 \times 10^{-6} \times \Delta T \][/tex]
[tex]\[ 0.1 = 9.9 \times 12 \times 10^{-6} \times \Delta T \][/tex]
[tex]\[ \Delta T = \frac{0.1}{9.9 \times 12 \times 10^{-6}} \][/tex]
[tex]\[ \Delta T \approx 841.75 \text{°C} \][/tex]
So, the steel ring must be heated to approximately 841.75°C above the current temperature.
### 8. Suppose a 300 L steel barrel is filled with Ethyl alcohol at a temperature of 10°C. How much Ethyl alcohol overflows if the system is heated to 100 °C?
To solve this, we need to consider:
1. The coefficient of volume expansion of Ethyl alcohol (approx. [tex]\( 7.5 \times 10^{-4} / \text{°C} \)[/tex]).
2. The coefficient of linear expansion of steel (because the volume expansion is three times the linear expansion, [tex]\( 3 \times \alpha \approx 36 \times 10^{-6} / \text{°C} \)[/tex]).
First, calculate the volumetric expansion of the steel barrel:
[tex]\[ \Delta V_{\text{steel}} = V_0 \times 3 \alpha \times \Delta T \][/tex]
[tex]\[ \Delta V_{\text{steel}} = 300 L \times 36 \times 10^{-6} / \text{°C} \times 90 \text{°C} \][/tex]
[tex]\[ \Delta V_{\text{steel}} = 0.972 L \][/tex]
Next, calculate the volumetric expansion of Ethyl alcohol:
[tex]\[ \Delta V_{\text{alcohol}} = V_0 \times \beta \times \Delta T \][/tex]
[tex]\[ \Delta V_{\text{alcohol}} = 300 L \times 7.5 \times 10^{-4} / \text{°C} \times 90 \text{°C} \][/tex]
[tex]\[ \Delta V_{\text{alcohol}} = 20.25 L \][/tex]
The overflow volume is the difference between the expansion of Ethyl alcohol and the barrel:
[tex]\[ V_{\text{overflow}} = \Delta V_{\text{alcohol}} - \Delta V_{\text{steel}} \][/tex]
[tex]\[ V_{\text{overflow}} = 20.25 L - 0.972 L \][/tex]
[tex]\[ V_{\text{overflow}} = 19.278 L \][/tex]
So, approximately 19.278 liters of Ethyl alcohol would overflow if the system is heated from 10°C to 100°C.
### 5. Why do we use mercury in a glass thermometer while the glass undergoes thermal expansion with it?
We use mercury in a glass thermometer for several reasons:
1. Wide Temperature Range: Mercury can remain in liquid form over a wide range of temperatures, from -39°C to 356°C. This makes it ideal for measuring a vast range of temperatures.
2. High Thermal Conductivity: Mercury conducts heat well, meaning it can quickly respond to changes in temperature.
3. Minimal Adhesion: Mercury does not wet glass; it forms a cohesive droplet within the glass tube. This prevents it from sticking to the walls, ensuring consistent and accurate readings.
4. Thermal Expansion: Though glass also undergoes thermal expansion, the expansion of mercury is significantly more pronounced. This means that even though glass expands, the volumetric expansion of mercury is the dominant factor influencing the reading.
5. Visibility: Mercury has a shiny, metallic appearance, making it easily visible within the glass tube.
### 6. When a metal ring and a metal sphere with equal radius are both at room temperature, the sphere does not pass through the ring. After the ring is heated, the sphere can pass through the ring. Why?
When the metal ring is heated, it undergoes thermal expansion. This means its dimensions increase, including its inner diameter. As the ring expands, the gap in the middle becomes larger, allowing the sphere with the same radius to pass through. The increase in the ring’s inner diameter due to heating compensates for the sphere’s radius that initially prevented it from passing through.
### 1. The diameter of a piston is 10 cm. To what temperature should a steel ring of diameter 9.9 cm be heated so that the piston can pass through the ring?
To solve this problem, we must consider the thermal expansion formula for solids:
[tex]\[ \Delta L = L_0 \alpha \Delta T \][/tex]
- [tex]\( \Delta L \)[/tex] is the change in length (diameter in this case).
- [tex]\( L_0 \)[/tex] is the original length (9.9 cm).
- [tex]\( \alpha \)[/tex] is the coefficient of linear expansion for steel (approx. [tex]\( 12 \times 10^{-6} / \text{°C} \)[/tex]).
- [tex]\( \Delta T \)[/tex] is the change in temperature.
For the diameter of the ring to increase from 9.9 cm to 10 cm:
[tex]\[ 10 - 9.9 = 9.9 \times 12 \times 10^{-6} \times \Delta T \][/tex]
[tex]\[ 0.1 = 9.9 \times 12 \times 10^{-6} \times \Delta T \][/tex]
[tex]\[ \Delta T = \frac{0.1}{9.9 \times 12 \times 10^{-6}} \][/tex]
[tex]\[ \Delta T \approx 841.75 \text{°C} \][/tex]
So, the steel ring must be heated to approximately 841.75°C above the current temperature.
### 8. Suppose a 300 L steel barrel is filled with Ethyl alcohol at a temperature of 10°C. How much Ethyl alcohol overflows if the system is heated to 100 °C?
To solve this, we need to consider:
1. The coefficient of volume expansion of Ethyl alcohol (approx. [tex]\( 7.5 \times 10^{-4} / \text{°C} \)[/tex]).
2. The coefficient of linear expansion of steel (because the volume expansion is three times the linear expansion, [tex]\( 3 \times \alpha \approx 36 \times 10^{-6} / \text{°C} \)[/tex]).
First, calculate the volumetric expansion of the steel barrel:
[tex]\[ \Delta V_{\text{steel}} = V_0 \times 3 \alpha \times \Delta T \][/tex]
[tex]\[ \Delta V_{\text{steel}} = 300 L \times 36 \times 10^{-6} / \text{°C} \times 90 \text{°C} \][/tex]
[tex]\[ \Delta V_{\text{steel}} = 0.972 L \][/tex]
Next, calculate the volumetric expansion of Ethyl alcohol:
[tex]\[ \Delta V_{\text{alcohol}} = V_0 \times \beta \times \Delta T \][/tex]
[tex]\[ \Delta V_{\text{alcohol}} = 300 L \times 7.5 \times 10^{-4} / \text{°C} \times 90 \text{°C} \][/tex]
[tex]\[ \Delta V_{\text{alcohol}} = 20.25 L \][/tex]
The overflow volume is the difference between the expansion of Ethyl alcohol and the barrel:
[tex]\[ V_{\text{overflow}} = \Delta V_{\text{alcohol}} - \Delta V_{\text{steel}} \][/tex]
[tex]\[ V_{\text{overflow}} = 20.25 L - 0.972 L \][/tex]
[tex]\[ V_{\text{overflow}} = 19.278 L \][/tex]
So, approximately 19.278 liters of Ethyl alcohol would overflow if the system is heated from 10°C to 100°C.
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