Homework Answers
![a) The rate at which the energy radiates from the person is, P= AEOT. = (1.55m”)(0.915)(5.6704x108W/m²K4)[(273.15 +32)K]* =](http://img.homeworklib.com/questions/1e7bb010-7296-11ea-846b-ff83fdc869c4.png?x-oss-process=image/resize,w_560)
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10/700 Resources Feedback The skin temperature of a nude person is 32.0°C and the surroundings are...
The skin temperature of a nude person is 32.0 °C and the surroundings are at 21.5°C....
The skin temperature of a nude person is 32.0 °C and the surroundings are at 21.5°C. The emissivity of skin is 0.900, and the surface area of the person is 1.55 m2 What is the rate P at which energy radiates from the person? 89 P = W What is the net energy loss AE from the body in 4 min by radiation? 21480 ΔΕ- J
Part A please thanks! The skin temperature of a nude person is 32.5°C and the surroundings...
Part A please thanks!
The skin temperature of a nude person is 32.5°C and the surroundings are at 21.0°C. The emissivity of skin is 0.910, and the surface area of the person is 1.50 m². What is the rate P at which energy radiates from the person? P = 507 What is the net energy loss AE from the body in 1 min by radiation? AE = 5755.2
Assuming your skin temperature is 37.2 degree C and the temperature of your surroundings is 21.8...
Assuming your skin temperature is 37.2 degree C and the temperature of your surroundings is 21.8 degree C. determine the length of time required for you to radiate away the energy gained by eating a 309-C ice cream cone. Let the emissivity of your skin be 0.915 and its area be 1.40 m^2.
How much power (power is the energy per second) is radiated by the person
Radiation of Energy The rate of heat transfer by emitted radiation is determined by the Stefan-Boltzmann law of radiation: = aeAT4 where o 5.67x10-8 J/s - m2 K is the Stefan-Boltzmann constant, A is the surface area of the object, and T is its absolute temperature in kelvin. The symbol e stands for the emissivity of the object, which is a measure of how well it radiates An ideal jet-black (or black body) radiator has e 1,whereas a perfect reflector has...
Estimate the lowest environment temperature that a person, who has skin temperature 34.0 °C, total skin...
Estimate the lowest environment temperature that a person, who has skin temperature 34.0 °C, total skin area 1.53 m2, emissivity 0.800, and metabolic energy production 157 J/s, can stand naked without a significant drop of body temperature.
The emissivity of the human skin is 97.0 percent. Use 35.0 °C for the skin temperature...
The emissivity of the human skin is 97.0 percent. Use 35.0 °C for the skin temperature and approximate the human body by a rectangular block with a height of 1.81 m, a width of 40.5 cm and a length of 32.0 cm. Calculate the power emitted by the human body. 1.430x103 w You are correct. Your receipt no. is 158-4715 Previous Tries Fortunately our environment radiates too. The human body absorbs this radiation with an absorbance of 97.0 percent, so...
A. The emissivity of the human skin is 97.0 percent. Use 35.0 °C for the skin...
A. The emissivity of the human skin is 97.0 percent. Use 35.0 °C for the skin temperature and approximate the human body by a rectangular block with a height of 1.55 m, a width of 33.0 cm and a length of 30.0 cm. Calculate the power emitted by the human body. B. What is the wavelength of the peak in the spectral distribution for this temperature? C. Fortunately our environment radiates too. The human body absorbs this radiation with an...
The total surface area of the human body is 1.20 m2 and the surface temperature is...
The total surface area of the human body is 1.20 m2 and the surface temperature is 30∘C=303∘K. If the surroundings are at a temperature of 6.0 ∘C , what is the net rate of heat loss from the body by radiation? The emissivity of the body is very close to unity, irrespective of skin pigmentation. Express your answer using two significant figures.
Calculate the rate of energy loss due to thermal radiation from an unclothed person if the average skin temperature is...
Calculate the rate of energy loss due to thermal radiation from an unclothed person if the average skin temperature is 37 °C and the room temperature is 20 °C. You may sume a total surface area of 1.5 m and a skin emissivity value of 0.9. How much energy (from food) must be consumed per day to compensate for this loss? If the person is surrounded by a layer of insulation of thickness 1.0 cm with a thermal conductivity of...
If your skin temperature increases from 30 ̊C to 37 ̊C, whilst in surroundings at temperature...
If your skin temperature increases from 30 ̊C to 37 ̊C, whilst in surroundings at temperature 15 ̊C, what happens to the rate of energy transfer from your skin to the surroundings a) by convection and b) by thermal radiation?
The skin temperature of a nude person is 32.0 °C and the surroundings are at 21.5°C. The emissivity of skin is 0.900, and the surface area of the person is 1.55 m2 What is the rate P at which energy radiates from the person? 89 P = W What is the net energy loss AE from the body in 4 min by radiation? 21480 ΔΕ- J
Part A please thanks!
The skin temperature of a nude person is 32.5°C and the surroundings are at 21.0°C. The emissivity of skin is 0.910, and the surface area of the person is 1.50 m². What is the rate P at which energy radiates from the person? P = 507 What is the net energy loss AE from the body in 1 min by radiation? AE = 5755.2
Assuming your skin temperature is 37.2 degree C and the temperature of your surroundings is 21.8 degree C. determine the length of time required for you to radiate away the energy gained by eating a 309-C ice cream cone. Let the emissivity of your skin be 0.915 and its area be 1.40 m^2.
Radiation of Energy The rate of heat transfer by emitted radiation is determined by the Stefan-Boltzmann law of radiation: = aeAT4 where o 5.67x10-8 J/s - m2 K is the Stefan-Boltzmann constant, A is the surface area of the object, and T is its absolute temperature in kelvin. The symbol e stands for the emissivity of the object, which is a measure of how well it radiates An ideal jet-black (or black body) radiator has e 1,whereas a perfect reflector has...
The emissivity of the human skin is 97.0 percent. Use 35.0 °C for the skin temperature and approximate the human body by a rectangular block with a height of 1.81 m, a width of 40.5 cm and a length of 32.0 cm. Calculate the power emitted by the human body. 1.430x103 w You are correct. Your receipt no. is 158-4715 Previous Tries Fortunately our environment radiates too. The human body absorbs this radiation with an absorbance of 97.0 percent, so...
Calculate the rate of energy loss due to thermal radiation from an unclothed person if the average skin temperature is 37 °C and the room temperature is 20 °C. You may sume a total surface area of 1.5 m and a skin emissivity value of 0.9. How much energy (from food) must be consumed per day to compensate for this loss? If the person is surrounded by a layer of insulation of thickness 1.0 cm with a thermal conductivity of...
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