# Find heat loss given metabolic rate in a closed system

1. Sep 16, 2014

### fishes

1. The problem statement, all variables and given/known data

A 60 kg person is exercising in the gym, doing external work at a rate of 200W. If they have an efficiency of 20%, calculate the rate of temperature increase of their body if none of this heat was able to be transferred to their surrounds.
(a) 828 ◦C per hour. (b) 13.8 ◦C per hour. (c) 3.45 ◦C per hour. (d) 0.86 ◦C per hour.

2. Relevant equations

Specific Heat of human = 3.473

3. The attempt at a solution

Really basic question but I just can't wrap my head around it. One thing that confuses me is that I thought the 20% efficiency meant that 20% was used for work and that 80% was lost as heat but the last part of the question says "if none of this heat"?
I'm guessing that the answer is 13.8 because that makes the most sense -here is my attempt so far:

I multiplied specific heat and the man's mass together to get 208380J/◦C. I then worked out what 200 watts was in hours = 720,000J/H. I then found both 20% and 80% of this value (144000J/H and 576000J/h respectively) as I wasn't sure what part of the energy generated was heat. I'm not too sure on what to do here now, I know we need to get a value with temperature over hour. The only way I can see to do this by dividing one of the 2 J/H values with the J/temp value but none of those equations gives me an answer. Am I missing something really easy? Does anyone have an suggestions on how to improve with these unit conversion/problem solving questions?

Cheers

2. Sep 16, 2014

### fishes

I got 3.45! So pretty much the thing that caused me the most confusion was what is generating the heat. Am I correct in normally assuming that when they say a human body is 20% efficient that means that 80% of the total energy generated to do work is lost to the environment as heat? The way I solved this question was taking the 200w to be the heat energy that is normally lost to the environment and plugged that in to get 3.45. Is it the question's fault for putting in the line about efficiency even though we didn't use it in the equation or mine for misunderstanding something major? Is the only heat lost to the environment the 20% of energy used to do work? I would really appreciate it if someone would go through my mad ramblings and tell me what I'm not understanding.

Cheers

3. Sep 16, 2014

### BvU

Hi there,
converting food (chemical energy) to work at 20% efficiency means that for every kJ of energy in the food 200 J can be used to do work (lift a weight, etc.); the remaining 800 J is heat. That heat is normally given off to the environment (breath, sweat, radiating, convection etc.). What you are asked to calculate is how fast this guy's body temperature goes up if he can't shed off that heat.

So the 3.45 $^\circ$/h is NOT the right answer! That 200 W IS exiting the "system" (i.e. the guy).

Your first guess was a lot better. It shows that if you do a lot of work (200 W is quite hefty), you really need to be able to dispose of the waste heat, or you'll go into fever conditions within ten minutes!

4. Sep 16, 2014

### fishes

Well at least I was right the first time concerning what percentage = the heat lost. Makes a lot more sense in regards to the question as well as you'll have to use the 200w to find out what the 80% of heat lost is! I've just got to get better at working through these types of problems. Thanks a lot for the help.