Respuesta :
Answer:
(a) 0.0001426 m²/s
(b) 544.9 W/Mk
(c) 210020.89 W/m²K
Explanation:
Let
Ta= ambient Temperature = 20°C
Initial temperature of hot dog(Ti) =20°C
Temperature of water(Tw) = 94°C
skin temperature(Ts) = 88°C
Centre temperature(Tc) = 59°C
Given that radius of hot dog, r = Diameter/2
r = 2.2/2 = 1.1cm = 0.011m
L= 12.5cm = 0.125m
Density(p,rho) = 980kg/m³
Specific heat capacity(Cp) = 3900J/kg K
Heat gained by hot dog(Hd) =
mCp( Tw-Ti)
Where mass = density × volume
We assume hot dog is cylindrical
Hd = p× (πr^2L)× Cp × (Tw-Ti)
Hd = 980(π × (0.011)² × 0.125 ) × 3900 × (94-20)
Hd = 13440.76J
Now given
Q = K A(Ts-Tc)/L
K = (Q × L)/A(Ts-Tc)
K = (Q × L)/2π rL ×(Ts-Tc)
K = Q / 2π r ×(Ts-Tc)
But heat lost(Hl) by hot dog after two minutes = p× (πr^2L)× Cp × (Ts-Ti)
Hl = 980(π × (0.011)² × 0.125 ) × 3900 × (88-20)
Hlost = 12350.97J
Q = 13,440.76 - 12350.97
Q = 1089.79J
K = 1089.79 /2π ×0.011 × (88-59)
K = 1089.79 /2.00
K = 544.895 Wm
(a) diffusivity = heat conducted per seconds/(rho(p) × Cp)
diffusivity = 544.90/980 ×3900
diffusivity = 0.0001426 m²/s
(b) K ~= 544.90 W/mk
(c) the convection heat transfer coefficient = Q-Qa/ A(Ts-Ta)
Since heat was transfer from water to the hot dog
Q-Qa = heat transfer from water to the hot dog = 13440.76
Hence transfer coefficient = 13440.76/[2πrL(Tw-Ts)]
= 13440.76/0.6394
= 21,020.89 W/m²K