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.) Express your answer with the appropriate units. Enter positive value if the person gives off heat and negative value if the person gains heat.

a) 1040 W

b) -87.21 W (minus sign mean that the Heat is gained from the environment)

c) 1127.21 W

d) 0.0279 kg of water is evaporated per minute.

e) 1.12 bottles

Explanation:

a) How much heat per second is produced just by the act of jogging?

During jogging, the body produces energy at a rate of 1300 W (given in the question). And 80% of rhe energy is converted to heat.

Heat per second produced while jogging

= 80% × 1300 = 1040 W

b) How much net heat per second does the runner gain just from radiation if the air temperature is 40.0∘C(104∘F)?

Heat lost by radiation from a surface or body is given by the Stefan-Boltzmann Law

Q = Aεσ (Tb⁴ - Ts⁴)

A = Surface area of the body available to lose heat by radiation = 1.85 m²

ε = emissivity of the body = 1 (logical assumption for this question)

σ = Stefan-Boltzmann constant = (5.67 ×10⁻⁸) W/m².K⁴

Tb = Temperature of the body in Kelvin = 33°C = 306.15 K

Ts = Temperature of the surroundings = 40°C = 313.15 K

Q = (1.85 × 1 × 5.67 × 10⁻⁸ × (306.15⁴ - 313.15⁴) = -87.21 W

The negative sign indicates that 87.21 W is the amount of heat gained from the environment.

c) What is the total amount of excess heat this runner’s body must get rid of per second?

Total amount of heat the runner must get rid of = (Heat produced by the runner's body) + (Heat gained from the environment) = 1040 + 87.21 = 1127.21 W

d) How much water must his body evaporate every minute due to his activity? The heat of vaporization of water at body temperature is 2.42×10⁶ J/kg.

Power = (Energy/time)

Energy = Power × Time

Power = 1127.21 W

Time = 1 minute = 60 s

Energy = 1127.21 × 60 = 67,632.78 J

This energy is then used to vaporize water

Energy = mL

m = mass of water vaporized = ?

L = Heat of vaporization of water at body temperature = (2.42×10⁶) J/kg

m = (Energy) ÷ L

m = (67,632.78) ÷ (2.42×10⁶) = 0.0279 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.

0.0279 kg of water is lost in a minute.

The amount of heat lost in half an hour or 30 minutes = 0.0279 × 30 = 0.838 kg

1 kg of water = 1.00 L

0.838 kg of water = 0.838 L of water.

Number of 750 mL water bottles required to replace the 838 mL water lost = (838/750) = 1.12 bottles of 750 mL water bottle is required.

Hope this Helps!!!

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