/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 25 A person eats a dessert that con... [FREE SOLUTION] | 91Ó°ÊÓ

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A person eats a dessert that contains 260 Calories. (This "Calorie" unit, with a capital \(\mathrm{C}\), is the one used by nutritionists; 1 Calorie \(=\) 4186 J. See Section 12.7.) The skin temperature of this individual is \(36^{\circ} \mathrm{C}\) and that of her environment is \(21{ }^{\circ} \mathrm{C}\). The emissivity of her skin is 0.75 and its surface area is \(1.3 \mathrm{m}^{2} .\) How much time would it take for her to emit a net radiant energy from her body that is equal to the energy contained in this dessert?

Short Answer

Expert verified
Approximately 2.13 hours.

Step by step solution

01

Understand the Problem

We need to determine how long it takes for a person to emit radiant energy equal to the energy from a dessert (260 Calories). 1 Calorie = 4186 J, so convert Calories to Joules.
02

Convert Calories to Joules

Calculate the energy in Joules: \[ 260 \text{ Calories} \times 4186 \text{ J/Calorie} = 1,088,360 \text{ J} \]
03

Use the Stefan-Boltzmann Law

The Stefan-Boltzmann law gives the power radiated by a body: \[ P = \varepsilon \sigma A (T_{\text{skin}}^4 - T_{\text{env}}^4) \]where \( \varepsilon = 0.75 \), \( \sigma = 5.67 \times 10^{-8} \text{ W/m}^2\text{K}^4 \), \( A = 1.3 \text{ m}^2 \), \( T_{\text{skin}} = 36 + 273 = 309 \text{ K} \), and \( T_{\text{env}} = 21 + 273 = 294 \text{ K} \).
04

Calculate Net Power Emitted

Insert values into the formula:\[ P = 0.75 \times 5.67 \times 10^{-8} \times 1.3 \times ((309)^4 - (294)^4) \]Calculate:\[ P \approx 0.75 \times 5.67 \times 10^{-8} \times 1.3 \times (9.1771 \times 10^8 - 7.4657 \times 10^8) \]\[ P \approx 142.14 \text{ W} \]
05

Calculate Time Required

The time \( t \) is given by the energy divided by the power:\[ t = \frac{E}{P} = \frac{1,088,360 \text{ J}}{142.14 \text{ W}} \approx 7657 \text{ s} \]
06

Convert Time to Hours

Convert seconds to hours:\[ t = \frac{7657}{3600} \approx 2.13 \text{ hours} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Radiant Energy Emission
Radiant energy emission refers to the energy an object gives off in the form of electromagnetic waves or radiation. This is a natural process through which all objects emit energy based on their temperature. The amount of radiation emitted increases with an object's temperature.

This emission of energy is significant because it's a fundamental way heat can be transferred without direct contact or a medium, unlike conduction and convection. If you're standing outside and feel warmth from the sun, that's radiant energy in action. It's responsible for the Earth receiving warmth from the sun. For objects on Earth, it means that, for example, humans or animals can radiate heat to their surroundings.

In physics and many practical situations, this concept is vital, as it helps in understanding how things cool down or warm-up. Thus, by looking at how much energy is emitted, we can predict temperature changes over time.
Thermal Radiation
Thermal radiation is the emission of electromagnetic waves from all matter with a temperature above absolute zero. This radiation is due to the thermal motion of charged particles within materials. With increased temperature, these particles move more vigorously, resulting in higher energy emissions.

Unique to thermal radiation is that it spans a range of the electromagnetic spectrum, from infrared to visible light, depending on the energy and temperature. At everyday temperatures, most thermal radiation is in the infrared range and is not visible to the human eye. However, at higher temperatures, such as those of the sun, it becomes visible.

A real-world example of thermal radiation is the warmth felt from a heat source, such as a campfire or radiator. Thermal radiation is critical in many scientific fields, including climate science and astrophysics, as well as in everyday technologies, like thermal imaging cameras.
Emissivity
Emissivity is a measure of a material's ability to emit thermal radiation compared to a perfect emitter, like a black body. The value of emissivity ranges from 0 to 1, where 1 signifies a perfect black body that emits all energy it receives, and 0 signifies a perfect reflector that emits none.

Understanding emissivity is essential when calculating radiant energy emissions in real-life objects, where it's rare to encounter perfect emitters. For example, a human body typically has an emissivity around 0.75. This means humans are quite effective at radiating heat, which impacts things like thermal comfort or energy loss.

Emissivity influences the Stefan-Boltzmann law, which is used to calculate the power radiated from a body. In practical situations, knowing the emissivity of materials can help in designing systems for heating or cooling and optimizing thermal regulation techniques. Thus, emissivity is a key factor in various applications, from engineering to the natural sciences.

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Most popular questions from this chapter

Two pots are identical except that the flat bottom of one is aluminum, whereas that of the other is copper. Water in these pots is boiling away at \(100.0^{\circ} \mathrm{C}\) at the same rate. The temperature of the heating element on which the aluminum bottom is sitting is \(155.0^{\circ} \mathrm{C} .\) Assume that heat enters the water only through the bottoms of the pots and find the temperature of the heating element on which the copper bottom rests.

A wood-burning stove (emissivity \(=0.900\) and surface area \(=3.50 \mathrm{m}^{2}\) ) is being used to heat a room. The fire keeps the stove surface at a constant \(198^{\circ} \mathrm{C}(471 \mathrm{K})\) and the room at a constant \(29^{\circ} \mathrm{C}(302 \mathrm{K}) .\) Determine the net radiant power generated by the stove.

The block in the drawing has dimensions \(L_{0} \times 2 L_{0} \times 3 L_{0},\) where \(L_{0}=0.30 \mathrm{m} .\) The block has a thermal conductivity of \(250 \mathrm{J} /\left(\mathrm{s} \cdot \mathrm{m} \cdot \mathrm{C}^{\circ}\right) .\) In drawings \(\mathrm{A}, \mathrm{B},\) and \(\mathrm{C},\) heat is conducted through the block in three different directions; in each case the temperature of the warmer surface is \(35^{\circ} \mathrm{C}\) and that of the cooler surface is \(19^{\circ} \mathrm{C} .\) Determine the heat that flows in \(5.0 \mathrm{s}\) for each case.

Liquid helium is stored at its boiling-point temperature of \(4.2 \mathrm{K}\) in a spherical container \((r=0.30 \mathrm{m})\). The container is a perfect blackbody radiator. The container is surrounded by a spherical shield whose temperature is \(77 \mathrm{K}\). A vacuum exists in the space between the container and the shield. The latent heat of vaporization for helium is \(2.1 \times 10^{4} \mathrm{J} / \mathrm{kg} .\) What mass of liquid helium boils away through a venting valve in one hour?

A baking dish is removed from a hot oven and placed on a cooling rack. As the dish cools down to \(35^{\circ} \mathrm{C}\) from \(175^{\circ} \mathrm{C}\), its net radiant power decreases to \(12.0 \mathrm{W}\). What was the net radiant power of the baking dish when it was first removed from the oven? Assume that the temperature in the kitchen remains at \(22^{\circ} \mathrm{C}\) as the dish cools down.

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