A person of A=1.8m^2 stands in a room T=295K for 24.0 hours. His skin temperatur

Question

A person of A=1.8m^2 stands in a room T=295K for 24.0 hours. His skin temperature remains a constant 306K and hisemissivity is 0.9. a)What is the net rate of radiative thermal transfer for theperson? Gain or Loss? b)what is the net thermal energy radiated by the person duringthe 24.0 hours? c) During this time, the person swallows a 0.30 kg sausage(T=275K, c=3450 J/kgºC). How much heat does the sausage gainas it is warmed up to 37.0ºC in the digestive tract?

a)What is the net rate of radiative thermal transfer for theperson? Gain or Loss? b)what is the net thermal energy radiated by the person duringthe 24.0 hours? c) During this time, the person swallows a 0.30 kg sausage(T=275K, c=3450 J/kgºC). How much heat does the sausage gainas it is warmed up to 37.0ºC in the digestive tract?

Explanation / Answer

power emitted by th4e person                           P =A(T4-T'4) = stephen boltzmann constant = emissitivity A = surface area of the personbody T = temparature of skin    T' = room temparature                     pluggingin the given values in the above equaion we can solve forP Thermal energy radiated from te person                 E = P * 24hr * 3600 sec / hr Heat absorbed by a 0.30 kg sausage be Q            Q = m*c * T     m = mass of the suasage      c = specific heat of suasage      T = final to intial temparaturedifference of suasage pluggin in given values , we can solve for aboveequations .   

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