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
[tex]P = \frac{E}{t} \rightarrow E= Energy, t = time[/tex]
[tex]P = 62W = 62 \frac{J}{s}[/tex]
The rate of energy loss per day would then be,
[tex]P = 62\frac{J}{s} (\frac{86400s}{1day})[/tex]
[tex]P = 5356800 \frac{J}{day}[/tex]
That is to say that Energy in Jules per lost day is 5356800J
By definition we know that [tex]1KCal = 4.184*10^{6}J[/tex]
In this way the energy in Cal is,
[tex]E = 5356800J \frac{1KCal}{4.184*10^{6}J}[/tex]
[tex]E = 1279.694 KCal[/tex]
The number of kilocalories (food calories) must be 1279.694 KCal
