A father and son conducted the following simple experiment on a hot dog which measured 12.5 cm in length and 2.2 cm in diameter. They inserted one food thermometer into the midpoint of the hot dog and another one was placed just under the skin of the hot dog. The temperatures of the thermometers were monitored until both thermometers read 20°C, which is the ambient temperature. The hot dog was then placed in 94°C boiling water and after exactly 2 minutes they recorded the center temperature and the skin temperature of the hot dog to be 59°C and 88°C, respectively. Assuming the following properties for the hot dog: rho = 980 kg/m3 and cp = 3900 J/kg·K and using transient temperature charts, determine (a) the thermal diffusivity of the hot dog, (b) the thermal conductivity of the hot dog, and (c) the convection heat transfer coefficient.

(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