21st Century Astronomy
21st Century Astronomy
6th Edition
ISBN: 9780393428063
Author: Kay
Publisher: NORTON
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Chapter 5, Problem 39QP
To determine

The power of the panel radiating into its surrounding.

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The temperature of the earth’s core is equal to 37270C, what is its equivalent in Kelvin scale ?
In a warm room a naked resting person has a skin temperature of 33°C if the room temperature is 29° C, what is the body surface area if the rate of heat loss due to convection is 43watt and the convection constant K =7.1 watt /m?.K Choose the right answer: 1.8m? 1.5m2 O 1.7m2
The following heat transfer formula quantifies the radiation emitted from the Sun: P = eoA(T4 – T?) Equation 5 where: P= radiated power (Watts) e = emissivity (=1 for ideal radiator; unitless) o = Stefan-Boltzmann constant = 5.67x10-8 W/m2-K+ A = radiating area (m²) T= temperature of radiator (Kelvin) Tc = temperature of surroundings (Kelvin) Q3 Using the following values, together with equation 5, calculate the power emitted by the Sun. sun's surface temperature = 5780 K temperature of the environment that the Sun is located in = 4 K emissivity of the Sun = 1 radius of the Sun = 695,700,000 m Stephen-Boltzmann constant o = 5.67 x 10-8 W/m2-K4 Show your work below-you may use the equation editor or insert a picture of your handwritten work.
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