$617 \textrm{ W}$

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# OpenStax College Physics, Chapter 14, Problem 9 (Problems & Exercises)

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*M*

*K*

*S*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

*Q*by multiplying both sides by

*T*, in which case we get

*Q*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

*T*out. We're going to do that here in red. And then, we'll divide the denominator

*T*to both terms in the numerator and we end up with power output required is the power input plus

*M*

*C*

*Delta T*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.

## Comments

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

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