## College Physics (4th Edition)

(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 $17.9~W$
(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 $17.9~W$