How much heat per second does a runner gain just from radiation?

[JustinLiang]

Homework Statement

When jogging strenuously, an average runner of mass 68kg and surface area 1.85m^2 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 33C instead of the usual 30C. (We shall neglect conduction) 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 40C? (Remember that he radiates out, but the environment radiates back in.)

Homework Equations

P=σeAT^4

The Attempt at a Solution

(a) This one is easy, just multiply 1300W by 0.8 which gets me 1040W.
(b) Given P=σeAT^4, I solved for the heat per second radiated into the body by the air (I am pretty sure emissivity is 1, it didn't specify...):
P=(5.6704x10^-8)(1)(1.85m^2)(273.15K+40K)^4
=1008.777W

Clearly this is wrong because 1008.777-1040 gives you a loss of heat? The answer suggests there is a gain of 87.1W.

What am I doing wrong?

Thanks.

 

[BruceW]

1040W is the total power output from the person, so this includes power output via sweating. Part b requires only the power output due to radiation, so you need to calculate this.

 

[JustinLiang]

1040W is the total power output from the person, so this includes power output via sweating. Part b requires only the power output due to radiation, so you need to calculate this.

In that case I get 1008W, but the answer is 87.1W?

 

[JustinLiang]

bump...

 

[rwisz]

As a scientist, it is important to carefully consider all factors when solving a problem. In this case, while the runner does radiate heat outwards, they also absorb heat from the environment. Therefore, it is necessary to account for both the radiated heat from the runner and the radiated heat from the environment in order to determine the net heat gained from radiation.

To do this, we can use the equation P = σeA(T^4-Tenv^4), where Tenv is the temperature of the environment in Kelvin. Plugging in the values given, we get:

P = (5.6704x10^-8)(1)(1.85m^2)((306.15K)^4-(313.15K)^4) = 87.1W

This is the net heat gained from radiation, taking into account both the runner's radiated heat and the heat absorbed from the environment. It is important to consider all factors and use the correct equations when solving scientific problems.