Question
Following vigorous exercise, the body temperature of an 80.0-kg person is $40.0^\circ\textrm{C}$. At what rate in watts must the person transfer thermal energy to reduce the the body temperature to $37.0^\circ\textrm{C}$ in 30.0 min, assuming the body continues to produce energy at the rate of 150 W? (1 watt = 1 joule / second or 1W = 1 J/s).

$617 \textrm{ W}$

Solution Video

# OpenStax College Physics for AP® Courses Solution, Chapter 14, Problem 9 (Problems & Exercises) (2:22) View sample solution

## Calculator Screenshots Video Transcript

This is College Physics Answers with Shaun Dychko. An 80-kilogram person has to cool their body from 40 degrees Celsius to 37 degrees Celsius in a time of 30 minutes. We convert that into 1800 seconds since all of our formulas require <i>M</i> <i>K</i> <i>S</i> units, meters, kilograms, and seconds. So, we take care of unit conversions when we're writing down the information that we know. The specific heat of a human body is 3500 Joules per kilogram per Celsius degree. And, we're told that the body is producing 150 watts, just near the metabolic rate at rest. And, the question is what rate of heat transfer out does the body require in order to reduce its temperature by three Celsius degrees in this time? So, this power output has to do two things. One, it has to account or compensate for the power being produced and then furthermore, it has to go beyond that to actually reduce temperature. So, in general, the power is energy divided by time. And, we can solve this for <i>Q</i> by multiplying both sides by <i>T</i>, in which case we get <i>Q</i> is power times time, and we'll use that here in a second. The power output of the body is going to be the heat output divided by time. The heat output is going to be the heat input, which is given to us at 150 watts multiplied by the time period. And then, we have to add to that the heat required to change the temperature by three Celsius degrees, 40 minus 37. And so, we'll substitute this in place of <i>T</i> out. We're going to do that here in red. And then, we'll divide the denominator <i>T</i> to both terms in the numerator and we end up with power output required is the power input plus <i>M</i> <i>C</i> <i>Delta T</i> over time. So, that's 150 watts being produced by the body plus 80 kilograms times the specific heat of the body multiplied by the change in temperature required divided by 1800 seconds. And, this means the power output has to be 617 watts.

Submitted by isabellaj. on Tue, 05/19/2020 - 19:33

Is the final temperature not 37 C? Delta T is final temperature minus initial temperature, correct? So Delta T will be -3 C?

Submitted by ShaunDychko on Wed, 05/20/2020 - 16:04

Thanks for the question isabellaj, you're quite right about Delta T being -3 C if one has a strict interpretation of "Q", however, I'm taking some liberties with negative signs here. Strictly speaking, Q is the heat gained by the body, and so using Delta T as -3 C correctly gives a negative Q since heat is in fact leaving the body. Nevertheless, I'm choosing to consider the magnitude of Q, and adding the rate of energy production of the body to the magnitude of the rate at which thermal energy needs to be transmitted away from the body. If you choose to have Delta T -3 C then you would also need to express the energy production by the body as negative, to get a total negative transmission of energy away from the body. The answers would have the same magnitude.
Hope this helps,
Shaun