Question

At room temperature, a typical person loses energy to the surroundings at the rate of 62 W. If this energy loss has to be made up by an equivalent food intake, how many kilocalories (food calories) does this person need to consume every day just to make up this heat loss

Answer 1

To solve this problem it is necessary to use the given proportions of power and energy, as well as the energy conversion factor in Jules to Calories.

The power is defined as the amount of energy lost per second and whose unit is Watt. Therefore the energy loss rate given in seconds was

$P = \frac{E}{t} \rightarrow E= Energy, t = time$

$P = 62W = 62 \frac{J}{s}$

The rate of energy loss per day would then be,

$P = 62\frac{J}{s} (\frac{86400s}{1day})$

$P = 5356800 \frac{J}{day}$

That is to say that Energy in Jules per lost day is 5356800J

By definition we know that $1KCal = 4.184*10^{6}J$

In this way the energy in Cal is,

$E = 5356800J \frac{1KCal}{4.184*10^{6}J}$

$E = 1279.694 KCal$

The number of kilocalories (food calories) must be 1279.694 KCal