Note: The diagram attached below is the completion of the given question.
Also calculate the heat loss from the coffee due to radiation and the total heat loss
Answer:
[tex]Q_{conv} = 16.04 W\\Q_{rad} = 3.20 W\\H_{loss} = 19.24 W[/tex]
Explanation:
Glass diameter of the coffee container, [tex]D_1 = 7 cm = 0.07 m[/tex]
Glass radius of the coffee container, r₁ = 0.07/2 = 0.035 m
Diameter of the aluminium housing, [tex]D_2 = 0.08 m[/tex]
Radius of the aluminium housing, r₂ = 0.08/2 = 0.04 m
Inner temperature, T₁ = 75°C = 348 K
Outer temperature, T₂ = 35°C = 308 K
[tex]\epsilon_1 = 0.25\\\epsilon_2 = 0.25[/tex]
[tex]T_f = \frac{T_1 + T_2}{2} \\T_f = \frac{348+308}{2} \\T_f = 328 K[/tex]
At [tex]T_f = 328 K[/tex], P = 1 atm
k = 0.0284 w/m-k, v = 23.74 * 10⁻⁶, [tex]\alpha = 26.6 * 10^{-6}[/tex], pr = 0.703, [tex]\beta = 3.05 * 10^{-3} K^{-1}[/tex]
[tex]H_{loss} = Q_{rad} + Q_{conv}[/tex]
[tex]Q_{conv} = \frac{2 \pi k L (T_1 - T_2)}{ln(\frac{r_2}{r_1} )} \\\\Q_{conv} = \frac{2 \pi* 0.0284 * 0.3 (75 - 35)}{ln(\frac{0.04}{0.035} )} \\Q_{conv} = 16.04 W[/tex]
[tex]Q_{rad} = \frac{\sigma (\pi D_1 L) (T_1^4 - T_2^4)}{\frac{1}{\epsilon_1 } + \frac{1 - \epsilon_2}{\epsilon_2} (\frac{\gamma_1}{\gamma_2}) }[/tex]
[tex]Q_{rad} = \frac{5.67*10^{-8}(\pi * 0.07*0.3) (348^4 - 308^4)}{\frac{1}{0.25 } + \frac{1 - 0.25}{0.25} (\frac{0.035}{0.04}) } \\\\Q_{rad} = 3.20 W[/tex]
[tex]H_{loss} = 3.20 + 16.04\\H_{loss} = 19.24 W[/tex]
