/*! 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 \(C,\) 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
It takes approximately 2.52 hours to emit the same energy contained in the dessert.

Step by step solution

01

Convert Calories to Joules

First, we need to convert the energy content of the dessert from Calories to Joules. The conversion factor is given: 1 Calorie = 4186 J. Therefore, 260 Calories can be converted to Joules by multiplying:\[ 260 \text{ Calories} \times 4186 \text{ J/Calorie} = 1,088,360 \text{ J} \]
02

Calculate Net Radiant Power Using Stefan-Boltzmann Law

The net radiant power emitted by the body can be calculated using the Stefan-Boltzmann Law:\[ P = \varepsilon \sigma A (T^4_{skin} - T^4_{env}) \]where \(\varepsilon\) is the emissivity (0.75), \(A\) is the surface area (1.3 \text{ m}^2), \(\sigma\) is the Stefan-Boltzmann constant \((5.67 \times 10^{-8} \text{ W/m}^2\text{K}^4)\), \(T_{skin}\) is the skin temperature, and \(T_{env}\) is the environment temperature. Convert the temperatures from Celsius to Kelvin:\[ T_{skin} = 36 + 273 = 309 \text{ K} \]\[ T_{env} = 21 + 273 = 294 \text{ K} \] Substitute these values into the Stefan-Boltzmann equation: \[ P = 0.75 \times 5.67 \times 10^{-8} \times 1.3 \times ((309)^4 - (294)^4) \]
03

Calculate Radiant Power

Perform the calculations from Step 2:- Compute \( 309^4 = 9.14769 \times 10^9 \text{ K}^4\)- Compute \( 294^4 = 7.45371 \times 10^9 \text{ K}^4\)- Subtract \( 9.14769 \times 10^9 - 7.45371 \times 10^9 = 1.69398 \times 10^9 \)Now calculate the net power:\[ P = 0.75 \times 5.67 \times 10^{-8} \times 1.3 \times 1.69398 \times 10^9 = 119.84 \text{ W} \]
04

Determine Time to Emit Radiant Energy

With the net energy in Joules and the power in Watts, we can find the time by dividing energy by power:\[ t = \frac{1,088,360 \text{ J}}{119.84 \text{ W}} = 9,080 \text{ s} \]Convert this time from seconds to hours:\[ 9,080 \div 3600 \approx 2.52 \text{ hours} \]

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Key Concepts

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

Stefan-Boltzmann Law
The Stefan-Boltzmann Law is an essential principle in determining how much energy an object emits in the form of radiation. It's particularly useful in understanding heat transfer without the need for direct contact. This law states that the power radiated by a black body is directly proportional to the fourth power of its temperature. However, real objects aren't perfect black bodies, which is why emissivity, a measure of an object's efficiency in emitting energy compared to a black body, comes into play. The mathematical expression of the Stefan-Boltzmann Law is:\[ P = \varepsilon \sigma A (T_{\text{skin}}^4 - T_{\text{env}}^4) \]
  • \(P\) represents the radiant power emitted, measured in watts (W).
  • \(\varepsilon\) is the emissivity factor, a unit-less measure ranging between 0 and 1 (with 1 being a perfect emitter).
  • \(\sigma\) is the Stefan-Boltzmann constant, valued at approximately \(5.67 \times 10^{-8} \text{ W/m}^2\text{K}^4\).
  • \(A\) stands for the surface area from which the radiation is emitted, measured in square meters (\(m^2\)).
  • \(T_{\text{skin}}\) and \(T_{\text{env}}\) are the absolute temperatures (in Kelvin) of the emitting body and its environment, respectively.
This law provides a clear pathway to calculate the net radiant energy exchange between a human body and its environment, crucial for understanding heat loss in various conditions.
Calories to Joules Conversion
In energy calculations, particularly regarding food energy, it's important to understand how to convert between units like Calories and Joules. The term 'Calorie' with an uppercase 'C' refers to the kilocalorie, which is commonly used in nutrition to describe the energy content in food. For precise calculations in physics and engineering, energy is usually expressed in Joules. The conversion between these units is straightforward:
  • 1 Calorie = 4186 Joules
Therefore, to convert from Calories to Joules, you simply multiply the number of Calories by 4186. For example, the energy content found in a dessert of 260 Calories translates to:\[ 260 \text{ Calories} \times 4186 \text{ J/Calorie} = 1,088,360 \text{ J} \]This conversion is a fundamental step in determining how much energy needs to be emitted by the body to match the intake from the dessert, a necessary calculation for the problem at hand.
Emissivity and Surface Area
Emissivity and surface area are critical factors in calculating radiant energy emission through the Stefan-Boltzmann Law.**Emissivity**:This value indicates how effectively a material emits thermal radiation compared to an ideal black body. With emissivity values ranging between 0 and 1:
  • An emissivity of 1 signifies perfect emission, typical for an ideal black body.
  • An emissivity of 0 implies no emission.
For human skin, typical values range between 0.7 to 0.9, with the problem using 0.75, suggesting that human skin is a relatively efficient emitter of thermal radiation.**Surface Area**:The surface area is the portion of the body from which radiation occurs. In our example, it's given as 1.3 \(\text{m}^2\). A greater surface area allows more area for thermal energy emission, influencing the net power result significantly.In combining both factors, emissivity and surface area dictate the actual output power when substituted into the Stefan-Boltzmann equation, influencing how energetically and efficiently a body radiates heat compared to its environment.

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Most popular questions from this chapter

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.

The Kelvin temperature of an object is \(T_{1},\) and the object radiates a certain amount of energy per second. The Kelvin temperature of the object is then increased to \(T_{2},\) and the object radiates twice the energy per second that it radiated at the lower temperature. What is the ratio \(T_{2} / T_{1} ?\)

One half of a kilogram of liquid water at \(273 \mathrm{K}\left(0^{\circ} \mathrm{C}\right)\) is placed outside on a day when the temperature is \(261 \mathrm{K}\left(-12^{\circ} \mathrm{C}\right) .\) Assume that the heat is lost from the water only by means of radiation and that the emissivity of the radiating surface is \(0.60 .\) Consider two cases: when the surface area of the water is (a) \(0.035 \mathrm{m}^{2}\) (as it might be in a cup) and (b) \(1.5 \mathrm{m}^{2}\) (as it could be if the water were spilled out to form a thin sheet). Concepts: (i) In case (a) is the heat that must be removed to freeze the water less than, greater than, or the same as in case (b)? (ii) The loss of heat by radiation depends on the temperature of the radiating object. Does the temperature of the water change as it freezes? (iii) The water both loses and gains heat by radiation. How, then, can heat transfer by radiation lead to freezing of the water? (iv) Will it take longer for the water to freeze in case (a) when the area is smaller or in case (b) when the area is larger? Calculations: For each case, (a) and (b), determine the time it takes the water to freeze into ice at \(0^{\circ} \mathrm{C}\).

A solid cylinder is radiating power. It has a length that is ten times its radius. It is cut into a number of smaller cylinders, each of which has the same length. Each small cylinder has the same temperature as the original cylinder. The total radiant power emitted by the pieces is twice that emitted by the original cylinder. How many smaller cylinders are there?

Suppose the skin temperature of a naked person is \(34^{\circ} \mathrm{C}\) when the person is standing inside a room whose temperature is \(25^{\circ} \mathrm{C} .\) The skin area of the individual is \(1.5 \mathrm{m}^{2} .\) (a) Assuming the emissivity is 0.80, find the net loss of radiant power from the body. (b) Determine the number of food Calories of energy (1 food Calorie = \(4186 \mathrm{J}\) ) that are lost in one hour due to the net loss rate obtained in part (a). Metabolic conversion of food into energy replaces this loss.

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