Physics for Scientists and Engineers
6th Edition
ISBN: 9781429281843
Author: Tipler
Publisher: MAC HIGHER
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Chapter 19, Problem 52P
To determine
The time taken by room’s air to warm.
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If the air temperature isthe same as the temperature of your skin (about 30°C), your body cannotget rid of heat by transferring it to the air. In that case, it gets ridof the heat by evaporating water (sweat). During bicycling, a typical70 kg person’s body produces energy at a rate of about 500 W due tometabolism, 80% of which is converted to heat. (a) How many kilogramsof water must the person’s body evaporate in an hour to get ridof this heat? The heat of vaporization of water at body temperature is2.42 * 10^6 J/kg. (b) The evaporated water must, of course, be replenished,or the person will dehydrate. How many 750 mL bottles of watermust the bicyclist drink per hour to replenish the lost water? (Recall thatthe mass of a liter of water is 1.0 kg.)
Hi there! Can you help em figure this problem. I keep gettign 707 watts and that is not correct.
Seals may cool themselves by using thermal windows, patches on their bodies with much higher than average surface temperature. Suppose a seal has a 0.030 m2 thermal window at a temperature of 30∘C.
If the seal's surroundings are a frosty -15 ∘C, what is the net rate of energy loss by radiation? Assume an emissivity equal to that of a human.
Express the answer in watts.
Thank you for the help! It is appreicated.
A cheese ripening room which, has a total surface area of A = 25.0m2, has to ensure that
the inside temperature does not exceed 11.0°C at all times. If the maximum outside
temperature is 35.0°C and the cheese ripening room has an air conditioner which provides
125 kW of cooling, calculate the how thi ck the wooden walls would need to bein mm, in
order to ensure the inside temperature do es not go above 11.0°C. Assume that the thermal
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conductivity, k of wood is 0.0800
Assume that the entire cheese ripening room is
m.C
made of wood (including ceiling and floor) and ignore any losses through the door as it
remains closed most of the time and is well sealed. Given: P = KA(Tinside –Toutside )
L
Chapter 19 Solutions
Physics for Scientists and Engineers
Ch. 19 - Prob. 1PCh. 19 - Prob. 2PCh. 19 - Prob. 3PCh. 19 - Prob. 4PCh. 19 - Prob. 5PCh. 19 - Prob. 6PCh. 19 - Prob. 7PCh. 19 - Prob. 8PCh. 19 - Prob. 9PCh. 19 - Prob. 10P
Ch. 19 - Prob. 11PCh. 19 - Prob. 12PCh. 19 - Prob. 13PCh. 19 - Prob. 14PCh. 19 - Prob. 15PCh. 19 - Prob. 16PCh. 19 - Prob. 17PCh. 19 - Prob. 18PCh. 19 - Prob. 19PCh. 19 - Prob. 20PCh. 19 - Prob. 21PCh. 19 - Prob. 22PCh. 19 - Prob. 23PCh. 19 - Prob. 24PCh. 19 - Prob. 25PCh. 19 - Prob. 26PCh. 19 - Prob. 27PCh. 19 - Prob. 28PCh. 19 - Prob. 29PCh. 19 - Prob. 30PCh. 19 - Prob. 31PCh. 19 - Prob. 32PCh. 19 - Prob. 33PCh. 19 - Prob. 34PCh. 19 - Prob. 35PCh. 19 - Prob. 36PCh. 19 - Prob. 37PCh. 19 - Prob. 38PCh. 19 - Prob. 39PCh. 19 - Prob. 40PCh. 19 - Prob. 41PCh. 19 - Prob. 42PCh. 19 - Prob. 43PCh. 19 - Prob. 44PCh. 19 - Prob. 45PCh. 19 - Prob. 46PCh. 19 - Prob. 47PCh. 19 - Prob. 48PCh. 19 - Prob. 49PCh. 19 - Prob. 50PCh. 19 - Prob. 51PCh. 19 - Prob. 52PCh. 19 - Prob. 53PCh. 19 - Prob. 54PCh. 19 - Prob. 55PCh. 19 - Prob. 56PCh. 19 - Prob. 57PCh. 19 - Prob. 58PCh. 19 - Prob. 59PCh. 19 - Prob. 60PCh. 19 - Prob. 61PCh. 19 - Prob. 62PCh. 19 - Prob. 63PCh. 19 - Prob. 64PCh. 19 - Prob. 65PCh. 19 - Prob. 66PCh. 19 - Prob. 67PCh. 19 - Prob. 68PCh. 19 - Prob. 69PCh. 19 - Prob. 70PCh. 19 - Prob. 71PCh. 19 - Prob. 72PCh. 19 - Prob. 73PCh. 19 - Prob. 74PCh. 19 - Prob. 75PCh. 19 - Prob. 76PCh. 19 - Prob. 77PCh. 19 - Prob. 78PCh. 19 - Prob. 79PCh. 19 - Prob. 80PCh. 19 - Prob. 81PCh. 19 - Prob. 82PCh. 19 - Prob. 83PCh. 19 - Prob. 84PCh. 19 - Prob. 85PCh. 19 - Prob. 86PCh. 19 - Prob. 87PCh. 19 - Prob. 88PCh. 19 - Prob. 89P
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