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A 1.88 -m ( 6 ft 2 in. ) man has a mass of 81 kg, a body surface area of 1.8 m 2 , and a skin temperature of 35.7 ∘ C. Normally, 80 % of the food calories he consumes go to heat; the rest goes to mechanical energy. His emissivity e is 1 because his body radiates almost entirely nonvisible infrared energy, which is not affected by skin pigment. (Careful! His body at 35.7 ∘ C radiates...
Given human body core temperature=37C, skin surface temperature=33C, surface area of body=1.5m^2, and average thickness=1.0cm. Thermal conductivity is 0.2J/(s*m*degree). If one takes into account heat conduction through the skin as the only mechanism for energy dissipation from the body how long would it take to dissipate all the energy contained in a food consisting of 140 Calories?
Consider a person with a surface area of 2.00 m and a skin temperature of 33.0°C. The person is in a room at 20.0°C. What is the net rate of heat transfer by radiation? Hint: See Example 12.23. Submit Request Answer
A man with a mass of 110 kg decides to lose weight by climbing the stairs in his office building several times a day. With his current daily activities and food intake, his weight is stable, but he would like to lose 0.5 kg per week without reducing food intake. Data shows that a person must burn 8000 Cal to lose 1 kg of mass (1 Cal = 4.19 x 103 J). If each story of the office building has...
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...
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...
how do i solve this problem? Multiple-Concept Example 8 reviews the approach that is used in problems such as this. A person eats a dessert that contains 180 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 the person is 36 °C and that of her environment is 17 °C. The emissivity of her skin is 0.84 and its surface area is 1.1...
The surface area of an average naked human adult is 2.2 m^2 and the average skin temperature is 33 degrees C. The emissivity of human skin is 0.95. Determine the radiation heat transfer rate from the human to the walls of a room if they are at 22 degrees C. This is why naked person feels chilly at room temperature.
Thank you so much! Part A You have probably seen people jogging in extremely hot weather and wondered Why" As we shall see, there are good reasons nat to do thisl When jogging strenuously, an average runner of mass 69.0 kg ?nd surface aea 1 84 m2 produces energy at a How much heat per second is produced just by the act ot jogging? rate of up to 1310 W 800 % of which is converted to heat The jogger...
2. The average person has 1.4 m2 of skin at a skin temperature of roughly 305 K (90°F) Consider the average person to be an ideal radiator standing in a room at a temperature of 293 K (68°F) (a) Calculate the power (energy per uni i) radiated by the average person in the form of blackbody radiation; express your answer in erg s-1. What is the person's "wattage"? (1 W = 107 erg s-1) Compare this to a typical incandescent...