#### Answer

(a) $1.55\times 10^6~J$ of heat leave the body each day due to breathing.
(b) The heat loss due to breathing is about 17% of the total heat loss each day.
(c) The average rate of heat loss is $17.9~W$

#### Work Step by Step

(a) We can find the number of moles:
$PV = nRT$
$n = \frac{PV}{RT}$
$n = \frac{(1.01\times 10^5~Pa)(5.0~m^3)}{(8.314~J/mol~K)(293.15~K)}$
$n = 207.2~mol$
We can find the required heat in one hour:
$Q = c~n~\Delta T$
$Q = (20.8~J/mol~K)(207.2~mol)(15~K)$
$Q = 64,646.4~J$
We can find the total heat required per day:
$Q = (24)(64,646.4~J) = 1.55\times 10^6~J$
$1.55\times 10^6~J$ of heat leave the body each day due to breathing.
(b) We can express this amount of heat as a fraction of the total heat loss of the body:
$\frac{1.55\times 10^6~J}{9\times 10^6~J} = 0.17$
The heat loss due to breathing is about 17% of the total heat loss each day.
(c) We can find the average rate of heat loss due to breathing:
$P = \frac{Q}{t} = \frac{1.55\times 10^6~J}{(24)(3600~s)} = 17.9~W$
The average rate of heat loss is $17.9~W$