The amount of heat per second conducted from the blood capillaries beneath the skin to the surface is 250 j/s. the energy is transferred a distance of 2.1 × 10-3 m through a body whose surface area is 1.9 m2. assuming that the thermal conductivity is that of body fat, determine the temperature difference between the capillaries and the surface of the skin.

Heat flux (i) = k.a.temp variation/d
i = 250 J/s or 250w ; d = 2.1 x 10^-3m ; A = 1.9m^2 ; K (Body Fat) = 0.2W/m.Kelvin ; Temp. Var. (Delta T) = ?

-> 250 = 0.2 x 1.9 x deltaT/2.1 x 10^-3
-> 250 x 2.1 x 10^-3 = 0.38 x deltaT
-> 525 x 10^-3 = 38x10^-2 x deltaT
-> deltaT = 525 x 10^-3/ 38 x 10^-2
-> deltaT = 13.81 x 10^(-3 - (-2))
-> deltaT = 13.81 x 10^-1
-> deltaT = 1.381 K

Explanation:

We shall apply law of conservation of momentum .

total Initial momentum =

7.5 x 8.85 = 66.375 kg m / s

total mass = 7.5 + 61.5 = 69 kg , common velocity of ball and performer be v

Total momentum = 69 v

69 v  = 66.375

v = 66.375 / 69

= .962 m /s .