Новое решение

astrosander правка от 16 апр. 2025 г., 05:40

правка #14295 позже →

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		+	### Statement
		+
		+	$13.4.17.$ [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 $D$:
		+
		+	$$D = \sqrt{\frac{4 L m}{\pi \varepsilon t}}$$
		+
		+	Substituting the numerical values gives:
		+	$D = 1.85 \text{ m}$
		+
		+	#### Answer
		+
		+	$D = 1.85 \text{ m}$
		+
		+	#### Answer
		+
		+	[Insert a concise answer or boxed result]

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			### Statement

			$13.4.17.$ [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 $D$:

			$$D = \sqrt{\frac{4 L m}{\pi \varepsilon t}}$$

			Substituting the numerical values gives:
			$D = 1.85 \text{ m}$

			#### Answer

			$D = 1.85 \text{ m}$

			#### Answer

			[Insert a concise answer or boxed result]