Edits to “Problem”, “Solution”, “Answer”
en/13.4.17.md
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| @@ -1,11 +1,5 @@ | |||
| − | ### | ||
| + | ### Problem | ||
| − | $13.4.17.$ [Insert the problem statement] | ||
| − | |||
| − | ### Solution | ||
| − | |||
| − | ### Problem Statement | ||
| − | |||
| $13.4.17.$ A vessel containing 100 g of water is placed at the focus of a spherical mirror. What should be the diameter of the mirror so that the water boils in 1 minute, assuming the mirror is directed straight at the Sun? The energy flux density from the Sun is 0.14 W/cm². Neglect any losses. | |||
| ### Solution | |||
| To boil the water, the required amount of energy is the heat of vaporization: | |||
| $$Q = L \cdot m \tag{1}$$ | |||
| where \(L\) is the specific heat of vaporization of water. The power needed to deliver this energy in time \(t\) is: | |||
| $$P = \frac{Q}{t} = \frac{L m}{t} \tag{2}$$ | |||
| This power is provided by the Sun’s radiation via the mirror: | |||
| $$P = \varepsilon \cdot S \tag{3}$$ | |||
| where \(\varepsilon\) is the energy flux density given in the problem, and \(S\) is the cross-sectional area of the spherical mirror. If the diameter of the mirror is denoted by \(D\), then \(S = \frac{\pi D^2}{4}\). Equating equations (2) and (3): | |||
| $$\frac{L m}{t} = \varepsilon \cdot \frac{\pi D^2}{4}$$ | |||
| Solving for \(D\): | |||
| @@ -30,13 +24,8 @@Solution | |||
| $$D = \sqrt{\frac{4 L m}{\pi \varepsilon t}}$$ | |||
| − | Substituting the numerical values gives: | ||
| − | \(D = 1.85 \text{ m}\) | ||
| + | Substituting the numerical values gives: \(D = 1.85 \text{ m}\) | ||
| #### Answer | |||
| − | |||
| − | |||
| − | #### Answer | ||
| − | |||
| − | [Insert a concise answer or boxed result] | ||
| + | \(D = 1.85 \text{ m}\) | ||
| @@ -1,11 +1,5 @@ | |||
| ### |
### Problem | ||
| $13.4.17.$ [Insert the problem statement] | |||
| ### Solution | |||
| ### Problem Statement | |||
| $13.4.17.$ A vessel containing 100 g of water is placed at the focus of a spherical mirror. What should be the diameter of the mirror so that the water boils in 1 minute, assuming the mirror is directed straight at the Sun? The energy flux density from the Sun is 0.14 W/cm². Neglect any losses. | $13.4.17.$ A vessel containing 100 g of water is placed at the focus of a spherical mirror. What should be the diameter of the mirror so that the water boils in 1 minute, assuming the mirror is directed straight at the Sun? The energy flux density from the Sun is 0.14 W/cm². Neglect any losses. | ||
| ### Solution | ### Solution | ||
| To boil the water, the required amount of energy is the heat of vaporization: | To boil the water, the required amount of energy is the heat of vaporization: | ||
| $$Q = L \cdot m \tag{1}$$ | $$Q = L \cdot m \tag{1}$$ | ||
| where \(L\) is the specific heat of vaporization of water. The power needed to deliver this energy in time \(t\) is: | where \(L\) is the specific heat of vaporization of water. The power needed to deliver this energy in time \(t\) is: | ||
| $$P = \frac{Q}{t} = \frac{L m}{t} \tag{2}$$ | $$P = \frac{Q}{t} = \frac{L m}{t} \tag{2}$$ | ||
| This power is provided by the Sun’s radiation via the mirror: | This power is provided by the Sun’s radiation via the mirror: | ||
| $$P = \varepsilon \cdot S \tag{3}$$ | $$P = \varepsilon \cdot S \tag{3}$$ | ||
| where \(\varepsilon\) is the energy flux density given in the problem, and \(S\) is the cross-sectional area of the spherical mirror. If the diameter of the mirror is denoted by \(D\), then \(S = \frac{\pi D^2}{4}\). Equating equations (2) and (3): | where \(\varepsilon\) is the energy flux density given in the problem, and \(S\) is the cross-sectional area of the spherical mirror. If the diameter of the mirror is denoted by \(D\), then \(S = \frac{\pi D^2}{4}\). Equating equations (2) and (3): | ||
| $$\frac{L m}{t} = \varepsilon \cdot \frac{\pi D^2}{4}$$ | $$\frac{L m}{t} = \varepsilon \cdot \frac{\pi D^2}{4}$$ | ||
| Solving for \(D\): | Solving for \(D\): | ||
| @@ -30,13 +24,8 @@Solution | |||
| $$D = \sqrt{\frac{4 L m}{\pi \varepsilon t}}$$ | $$D = \sqrt{\frac{4 L m}{\pi \varepsilon t}}$$ | ||
| Substituting the numerical values gives: | Substituting the numerical values gives: \(D = 1.85 \text{ m}\) | ||
| \(D = 1.85 \text{ m}\) | |||
| #### Answer | #### Answer | ||
| \(D = 1.85 \text{ m}\) | |||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||