Rate body temperature increases given rate of heat transfer?

In summary, the task is to calculate the rate of heat transfer to a person in a sauna with an ambient temperature of 57.0°C, given their skin temperature is 37.0°C, emissivity of skin is 0.95, and body surface area is 1.60 m2. The solution for the first part is 226W. The next part is to determine the rate at which the person's body temperature will increase (in degC/s) if their mass is 68.0 kg, with the specific heat of the human body being 3500 J/kg*K. The relevant equations are Q = mC(ΔT) and Power = Energy/Time. By rearranging the equations,
  • #1
CrashMaverick
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Homework Statement


Suppose you walk into a sauna that has an ambient temperature of 57.0°C. Calculate the rate of heat transfer to you by radiation given your skin temperature is 37.0°C, the emissivity of skin is 0.95, and the surface area of your body is 1.60 m2.

I solved this first part and found the answer to be 226W

If all other forms of heat transfer are balanced (the net heat transfer is zero), at what rate will your body temperature increase (degC/s) if your mass is 68.0 kg?

Note from instructor:

You can take the specific heat of the human body as:

c = 3500 J/kg*K

Homework Equations



Q = mC(ΔT)? Not entirely sure where to start really

m = mass
C = specific heat
ΔT = change in temperature

The Attempt at a Solution


So if you multiple C by m and ΔT(in kelvin) you end up with units of Joules right?
1 joule / 1 sec = 1 watt, I figure there is something I can do there, but I'm not sure what. The thing that is throwing me off is the units of degrees Celsius per second.

I've been staring at this problem for a couple hours and I'm sure I'm overlooking a very simple thing, any help is appreciated.
 
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  • #2
Your relevant equation is relevant. The Q does indeed stand for energy (in Joules).

The other equation you need is Power = Energy/Time.

Put the two equations together and rearrange so that on one side you have the quantities temperature and time as those are the units specified for the answer.

Sorry for all the edits I made to this repy.
 

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