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?
Homework Answers
my dear student nothing is given at a place of A please put Area
in this equation you will definitely get the answerr
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If your skin temperature increases from 30 ̊C to 37 ̊C, whilst in surroundings at temperature...
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
10/700 Resources Feedback The skin temperature of a nude person is 32.0°C and the surroundings are...
10/700 Resources Feedback The skin temperature of a nude person is 32.0°C and the surroundings are at 23.0°C. The emissivity of skin is 0.915, and the surface area of the person is 1.55 m² What is the rate Pat which energy radiates from the person? P= 78.57 What is the net energy loss AE from the body in 3 min by radiation? AE = 14143
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.
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...
art II- Show your work 17 Marks team at 320°C flows in a stainless steel pipe...
art II- Show your work 17 Marks team at 320°C flows in a stainless steel pipe (kr 15 W/m "C) whose inner and outer diameters are 5 em and 5.5 cm, respectively. The pipe is covered with 3-cm-thick glass wool insulation (k 0.038 W m·°C). Heat is lost to the surroundings at 5°C by natural convection and radiation, with a combined natural convection and radiation heat transfer coefficient of hi-15 w/m.。C. The convective heat transfer coefficient inside the pipe to...
On a windy day, we can define the wind-chill temperature Twc as the air temperature that...
On a windy day, we can define the wind-chill temperature Twc as the air temperature that would give the same rate of heat loss if experienced on a calm day. Consider a day when the air temperature is 2°C and the wind speed is 12 m/s, giving a convective heat transfer coefficient of 45 W/m2⋅K (compared with 10 W/m2⋅K on a calm day). Assume a person has a 3 mm layer of skin (k = 0.37 W/m⋅K), a body surface...
The temperature within the Earth's crust increases about 1.0 C∘ for each 30 m of depth....
The temperature within the Earth's crust increases about 1.0 C∘ for each 30 m of depth. The thermal conductivity of the crust is 0.80 W/C∘⋅m. Part A Determine the heat transferred from the interior to the surface for the entire Earth in 9.0 h . Express your answer using two significant figures. Part B Compare this heat to the amount of energy incident on the Earth in 9.0 h due to radiation from the Sun. Express your answer using two...
A household oven door of 0.6-m height and 0.7-m width reaches an average surface temperature of...
A household oven door of 0.6-m height and 0.7-m width reaches an average surface temperature of 32°C during operation. The temperature of ambient air is 22°C. Assuming the door has an emissivity of 0.8 and the surroundings are also at 22°C, determine the heat transfer by free convection and the heat transfer by radiation. 1) What is the volumetric thermal expansion coefficient of air at the film temperature outside the oven door, in 1/K? 2) What is the Rayleigh number...
The temperature within the Earth's crust increases about 1.0 C∘ for each 30 m of depth....
The temperature within the Earth's crust increases about 1.0 C∘ for each 30 m of depth. The thermal conductivity of the crust is 0.80 W/C∘⋅m. a)Determine the heat transferred from the interior to the surface for the entire Earth in 7.0 h. Answer for this is : 3.4×1017 b)Compare this heat to the amount of energy incident on the Earth in 7.0 h due to radiation from the Sun. Am not sure how to do part B. Any help is...
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
10/700 Resources Feedback The skin temperature of a nude person is 32.0°C and the surroundings are at 23.0°C. The emissivity of skin is 0.915, and the surface area of the person is 1.55 m² What is the rate Pat which energy radiates from the person? P= 78.57 What is the net energy loss AE from the body in 3 min by radiation? AE = 14143
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.
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...
art II- Show your work 17 Marks team at 320°C flows in a stainless steel pipe (kr 15 W/m "C) whose inner and outer diameters are 5 em and 5.5 cm, respectively. The pipe is covered with 3-cm-thick glass wool insulation (k 0.038 W m·°C). Heat is lost to the surroundings at 5°C by natural convection and radiation, with a combined natural convection and radiation heat transfer coefficient of hi-15 w/m.。C. The convective heat transfer coefficient inside the pipe to...
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