infrared camera temperature measure
Physics thermal physics?
can someone fix this? Show me the process, the answers are already there, but I need to understand why .. (B) A chilly bin has walls 4.2 cm thick and the total area of the walls is 0.8 m2. The chilly bin is loaded with 4.0 kg of ice at 0 ° C and stood on a grid so that the entire surface in contact with the air. The temperature on the surface of the chilly bin is 35.0 ° C. If the chilly bin is made of polystyrene (kstyrofoam J = 0.01 m s-1-1 ° C-1), how many hours it will take all the ice melting? (Note: Lfwater = 3.35 × 105 J kg-1) 55.8 hours (c) Identifying persons who are suspected of being infected with SARS, airports around the world have set up an infrared camera that the infrared radiation that people give off measures. What is the percentage increase in the rate of radiant heat of a person with a surface of the skin temperature of 38.0 ° C compared to the same person with a skin temperature of 33 ° C? 6.7%
The external tank will heat conduction time that will melt the ice at the speed of heat gain by consuction in t-dot Q = dQ / T = k A [35-0] / tickness -------------- ----- --------------------- k thermal conductivity of bin, A is the area --------------- dQ = heat of melting ice at 0 C for 4 kg = 3.35 ml * dQ = 4 × 105 dQ / T = k A [35-0] / tickness t = thickness * dQ / k A [35-0] t = 0.042 * 4 * 3.35 × 10 ^ 5 / 0.01 * 0.8 * 35 = 0.042 T * 4 * 3.35 × 10 ^ 5 / 0.01 * 0.8 * 35 = t T = 201000/3600 201000 secs = 55,833 hours -------------------------- rate of heat loss by radiation Q-dot = dQ / dt A sigma = [T ^ 4] Almost sigma is Stefan constant, A surface temp in T = 0K Q1 dot sigma A [311 ^ 4]>>> for t = 38 ° C Q2 = dot-sigma A [306 ^ 4]>>> for t = 33 ° C ----- --------------------------------- Both have adopted the same person differently Q-dot/Area A comparison will lead to similar results ---------------------------- [Q1-dot - Q2-dot] 100 * / Q2-dot% =% increase increse = [311 ^ 4 to 306 ^ 4] / [306 ^ 4] = 6.6978%
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