Answer:[tex]Q=248.011 W[/tex]
Explanation:
Given
Temperature of Room [tex]T_{\infty }=18^{\circ}\approx 291 K[/tex]
Area of Person [tex]A=1.7 m^2[/tex]
Temperature of skin [tex]T=32^{\circ}\approx 305 K[/tex]
Heat transfer coefficient [tex]h=5 W/m^2.k[/tex]
Emissivity of the skin and clothes [tex]\epsilon =0.9[/tex]
[tex]\Delta T=32-18=14^{\circ}[/tex]
Total rate of heat transfer=heat Transfer due to Radiation +heat transfer through convection
Heat transfer due radiation [tex]Q_1=\epsilon \sigma A(T^4-T_{\infty }^4 )[/tex]
where [tex]\sigma =stefan-boltzman\ constant[/tex]
[tex]Q_1=0.9\times 5.687\times 10^{-8}\times 1.7\times (305^4-291^4) [/tex]
[tex]Q_1=129.01 W[/tex]
Heat Transfer due to convection is given by
[tex]Q_2=hA(\Delta T)[/tex]
[tex]Q_2=5\times 1.7\times 14=119 W[/tex]
[tex]Q=Q_1+Q_2[/tex]
[tex]Q=129.01+119=248.011 W[/tex]