Problem 17.109 You have probably seen people jogging in extremely hot weather an
ID: 1504246 • Letter: P
Question
Problem 17.109
You have probably seen people jogging in extremely hot weather and wondered "Why?" As we shall see, there are good reasons not to do this! When jogging strenuously, an average runner of mass 65.0 kg and surface area 1.87 m2 produces energy at a rate of up to 1320 W , 80.0 % 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 33.0 C instead of the usual 30.0 C. (We shall neglect 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).
Part A
How much heat per second is produced just by the act of jogging?
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Part 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 that he radiates out, but the environment radiates back in.)
Part C
What is the total amount of excess heat this runner's body must get rid of per second?
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Part D
How much water must the jogger's body evaporate every minute due to his activity? The heat of vaporization of water at body temperature is 2.42×106 J/kg .
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Part 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 1.00 kg .
Problem 17.109
You have probably seen people jogging in extremely hot weather and wondered "Why?" As we shall see, there are good reasons not to do this! When jogging strenuously, an average runner of mass 65.0 kg and surface area 1.87 m2 produces energy at a rate of up to 1320 W , 80.0 % 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 33.0 C instead of the usual 30.0 C. (We shall neglect 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).
Part A
How much heat per second is produced just by the act of jogging?
Pjog = J/sSubmitMy AnswersGive Up
Part 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 that he radiates out, but the environment radiates back in.)
Hnet = WPart C
What is the total amount of excess heat this runner's body must get rid of per second?
Htotal = J/sSubmitMy AnswersGive Up
Part D
How much water must the jogger's body evaporate every minute due to his activity? The heat of vaporization of water at body temperature is 2.42×106 J/kg .
m = gSubmitMy AnswersGive Up
Part 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 1.00 kg .
n = bottlesExplanation / Answer
heat per second is produced just by the act of jogging= 80% of energy produced = 1320*0.8=1056 J/s
b) J=(Ts4- T4 )= 5.67*10^-8*(313^4-306^4) =47.07 W
net heat per second does the runner gain just from radiation if the air temperature is 40.0 C (104 F) is Q=JA= 88.02 W
c)Htotal= Q+1056 =1144 J /s
d)the amount of m kg of water must the jogger's body evaporate every minute due to his activity
m*2.42*10^6=1144*60
m= 0.028 kg =28 gm
e) the weight of 1 L is 1 kg.
the amount of m kg of water must the jogger's body evaporate every half an hour due to his activity
.028*30=0.85 kg
needed water= 0.85/.75 =1.13 bottle