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en/5.5.8.md
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| ### Statement | |||
| − | $5.5.8.$ [Insert the problem statement] | ||
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| − | ### Solution | ||
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| − | ### Statement | ||
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| $5.5.8.$ The air pressure inside a bottle is $0.1 \, \text{MPa}$ at a temperature of $7 \, ^\circ\text{C}$. By how much must the bottle be heated for the cork to pop out? Without heating, the cork can be removed by applying a force of $10 \, \text{N}$ to it. The cross-section of the cork is $2 \, \text{cm}^2$. | |||
| ### Solution | |||
| Let the initial pressure inside the bottle be $p_1 = 0.1 \, \text{MPa} = 10^5 \, \text{Pa}$, which is equal to the external atmospheric pressure $p_0$. The initial temperature is $T_1 = 7 \, ^\circ\text{C} = 280 \, \text{K}$. | |||
| The force required to pull the cork out without heating is $F = 10 \, \text{N}$, and the cross-sectional area is $S = 2 \, \text{cm}^2 = 2 \cdot 10^{-4} \, \text{m}^2$. | |||
| When the bottle is heated, the cork will pop out when the pressure difference between the inside and outside creates an outward force equal to the restraining static friction force holding the cork in place. This frictional force is exactly equal to the force $F$ needed to pull it out initially. | |||
| The final pressure $p_2$ required to pop the cork satisfies the equilibrium condition: | |||
| $$(p_2 - p_0) S = F$$ | |||
| $$p_2 = \frac{F}{S} + p_0$$ | |||
| Since the volume of the bottle remains constant before the cork pops, we can use the isochoric process law for the air inside: | |||
| $$\frac{p_1}{T_1} = \frac{p_2}{T_2}$$ | |||
| Expressing the final temperature $T_2$: | |||
| $$T_2 = \frac{p_2 T_1}{p_1} = \frac{\left(\frac{F}{S} + p_0\right) T_1}{p_1}$$ | |||
| Let us calculate the value of the additional pressure created by the friction force: | |||
| $$\frac{F}{S} = \frac{10}{2 \cdot 10^{-4}} = 5 \cdot 10^4 \, \text{Pa} = 0.5 \cdot 10^5 \, \text{Pa}$$ | |||
| Now, substituting the values into the temperature equation: | |||
| $$T_2 = \frac{(0.5 \cdot 10^5 + 10^5) \cdot 280}{10^5} = 1.5 \cdot 280 = 420 \, \text{K}$$ | |||
| The required change in temperature is: | |||
| $$\Delta T = T_2 - T_1 = 420 \, \text{K} - 280 \, \text{K} = 140 \, \text{K}$$ | |||
| Since a temperature change in Kelvin is numerically equal to a temperature change in degrees Celsius, $\Delta t = 140 \, ^\circ\text{C}$. | |||
| @@ -39,7 +33,3 @@Solution | |||
| #### Answer | |||
| $\Delta t = 140 \, ^\circ\text{C}$ | |||
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| − | #### Answer | ||
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| − | [Insert a concise answer or boxed result] | ||
| @@ -1,11 +1,5 @@ | |||
| ### Statement | ### Statement | ||
| $5.5.8.$ [Insert the problem statement] | |||
| ### Solution | |||
| ### Statement | |||
| $5.5.8.$ The air pressure inside a bottle is $0.1 \, \text{MPa}$ at a temperature of $7 \, ^\circ\text{C}$. By how much must the bottle be heated for the cork to pop out? Without heating, the cork can be removed by applying a force of $10 \, \text{N}$ to it. The cross-section of the cork is $2 \, \text{cm}^2$. | $5.5.8.$ The air pressure inside a bottle is $0.1 \, \text{MPa}$ at a temperature of $7 \, ^\circ\text{C}$. By how much must the bottle be heated for the cork to pop out? Without heating, the cork can be removed by applying a force of $10 \, \text{N}$ to it. The cross-section of the cork is $2 \, \text{cm}^2$. | ||
| ### Solution | ### Solution | ||
| Let the initial pressure inside the bottle be $p_1 = 0.1 \, \text{MPa} = 10^5 \, \text{Pa}$, which is equal to the external atmospheric pressure $p_0$. The initial temperature is $T_1 = 7 \, ^\circ\text{C} = 280 \, \text{K}$. | Let the initial pressure inside the bottle be $p_1 = 0.1 \, \text{MPa} = 10^5 \, \text{Pa}$, which is equal to the external atmospheric pressure $p_0$. The initial temperature is $T_1 = 7 \, ^\circ\text{C} = 280 \, \text{K}$. | ||
| The force required to pull the cork out without heating is $F = 10 \, \text{N}$, and the cross-sectional area is $S = 2 \, \text{cm}^2 = 2 \cdot 10^{-4} \, \text{m}^2$. | The force required to pull the cork out without heating is $F = 10 \, \text{N}$, and the cross-sectional area is $S = 2 \, \text{cm}^2 = 2 \cdot 10^{-4} \, \text{m}^2$. | ||
| When the bottle is heated, the cork will pop out when the pressure difference between the inside and outside creates an outward force equal to the restraining static friction force holding the cork in place. This frictional force is exactly equal to the force $F$ needed to pull it out initially. | When the bottle is heated, the cork will pop out when the pressure difference between the inside and outside creates an outward force equal to the restraining static friction force holding the cork in place. This frictional force is exactly equal to the force $F$ needed to pull it out initially. | ||
| The final pressure $p_2$ required to pop the cork satisfies the equilibrium condition: | The final pressure $p_2$ required to pop the cork satisfies the equilibrium condition: | ||
| $$(p_2 - p_0) S = F$$ | $$(p_2 - p_0) S = F$$ | ||
| $$p_2 = \frac{F}{S} + p_0$$ | $$p_2 = \frac{F}{S} + p_0$$ | ||
| Since the volume of the bottle remains constant before the cork pops, we can use the isochoric process law for the air inside: | Since the volume of the bottle remains constant before the cork pops, we can use the isochoric process law for the air inside: | ||
| $$\frac{p_1}{T_1} = \frac{p_2}{T_2}$$ | $$\frac{p_1}{T_1} = \frac{p_2}{T_2}$$ | ||
| Expressing the final temperature $T_2$: | Expressing the final temperature $T_2$: | ||
| $$T_2 = \frac{p_2 T_1}{p_1} = \frac{\left(\frac{F}{S} + p_0\right) T_1}{p_1}$$ | $$T_2 = \frac{p_2 T_1}{p_1} = \frac{\left(\frac{F}{S} + p_0\right) T_1}{p_1}$$ | ||
| Let us calculate the value of the additional pressure created by the friction force: | Let us calculate the value of the additional pressure created by the friction force: | ||
| $$\frac{F}{S} = \frac{10}{2 \cdot 10^{-4}} = 5 \cdot 10^4 \, \text{Pa} = 0.5 \cdot 10^5 \, \text{Pa}$$ | $$\frac{F}{S} = \frac{10}{2 \cdot 10^{-4}} = 5 \cdot 10^4 \, \text{Pa} = 0.5 \cdot 10^5 \, \text{Pa}$$ | ||
| Now, substituting the values into the temperature equation: | Now, substituting the values into the temperature equation: | ||
| $$T_2 = \frac{(0.5 \cdot 10^5 + 10^5) \cdot 280}{10^5} = 1.5 \cdot 280 = 420 \, \text{K}$$ | $$T_2 = \frac{(0.5 \cdot 10^5 + 10^5) \cdot 280}{10^5} = 1.5 \cdot 280 = 420 \, \text{K}$$ | ||
| The required change in temperature is: | The required change in temperature is: | ||
| $$\Delta T = T_2 - T_1 = 420 \, \text{K} - 280 \, \text{K} = 140 \, \text{K}$$ | $$\Delta T = T_2 - T_1 = 420 \, \text{K} - 280 \, \text{K} = 140 \, \text{K}$$ | ||
| Since a temperature change in Kelvin is numerically equal to a temperature change in degrees Celsius, $\Delta t = 140 \, ^\circ\text{C}$. | Since a temperature change in Kelvin is numerically equal to a temperature change in degrees Celsius, $\Delta t = 140 \, ^\circ\text{C}$. | ||
| @@ -39,7 +33,3 @@Solution | |||
| #### Answer | #### Answer | ||
| $\Delta t = 140 \, ^\circ\text{C}$ | $\Delta t = 140 \, ^\circ\text{C}$ | ||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||