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Get Full Access to University Physics - 13 Edition - Chapter 17 - Problem 118p
Get Full Access to University Physics - 13 Edition - Chapter 17 - Problem 118p

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# Overheating While Jogging. (a) if The jogger in The

ISBN: 9780321675460 31

## Solution for problem 118P Chapter 17

University Physics | 13th Edition

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Problem 118P

Overheating While Jogging. (a) if The jogger in The preceding problem were not able to get rid of the excess heat, by how much would his body temperature increase above the normal 37°C in a half hour of jogging? The specific heat for a human is about 3500 J/kg · K. (b) How high a fever (in °F) would this temperature increase be equivalent to? Is the increase large enough to be of concern? (Recall that normal body temperature is 98.6°F.) PRECEDING PROBLEM: BIO Jogging in the Heat of the Day. You have probably seen people jogging in extremely hot weather. There are good reasons not to do this! When jogging strenuously, an average runner of mass 68 kg and surface area 1.85 m2 produces energy at a rate of up to 1300 W, 80% of which is converted to heat. The jogger radiates heat but actually absorbs more from the hot air than he radiates away. At such high levels of activity, the skin’s temperature can be elevated to around 33°C instead of the usual 30°C. (Ignore conduction, which would bring even more heat into his body.) The only way for the body to get rid of this extra heat is by evaporating water (sweating). (a) How much heat per second is produced just by the act of jogging? (b) How much net heat per second does the runner gain just from radiation if the air temperature is 40.0°C (104°F)? (Remember: He radiates out, but the environment radiates back in.) (c) What is the total amount of excess heat this runner’s body must get rid of per second? (d) How much water must his body evaporate every minute due to his activity? The heat of vaporization of water at body temperature is 2.42 X 106 J/kg. (e) How many 750-mL bottles of water must he drink after (or preferably before!) jogging for a half hour? Recall that a liter of water has a mass of 1.0 kg.

Step-by-Step Solution:

Solution 118P Step 1: In the preceding problem, the power of the person who is jogging, P = 1300 W The percentage of heat energy generated, Q = 80% Therefore, the amount of heat energy generated per second, P heat 1300 W (80/100) = 1040 W Step 2: The person is jogging for half an hour as per the previous problem. ½ hour = 30 minutes 1 minute = 60 s 30 minutes = 30 × 60 s = 1800 s Therefore, the amount of heat energy generated, Q = P heat t Q = 1040 W × 1800 s = 1872000 J

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