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Hot coffee is contained in a cylindrical thermos bottle that is of length L=0.3 m and is lying on its side (horizontally). The coffee container consists of a glass flask of diameter D1=0.07mD_{1}=0.07 \mathrm{m}, separated from an aluminum housing of diameter D2=0.08mD_{2}=0.08 \mathrm{m} by air at atmospheric pressure. The outer surface of the flask and the inner surface of the housing are silver coated to provide emissivities of ε1=ε2=0.25\varepsilon_{1}=\varepsilon_{2}=0.25. If these surface temperatures are T1=75CT_{1}=75^{\circ} \mathrm{C} and T2=35CT_{2}=35^{\circ} \mathrm{C}, what is the heat loss from the coffee?

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Answered 2 years ago
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We are given following data:\textbf{We are given following data:}

T1=75° C=75+273=348 KT2=35° C=35+273=308 KD1=0.07 mR1=D12=0.07m2=0.035 mD2=0.08 mR2=D22=0.08m2=0.04 mϵ1=ϵ2=0.25L=0.3 m\begin{align*} T_{1}&=75\text{\textdegree}\mathrm{~C}\\ &=75+273\\ &=348\mathrm{~K}\\ T_{2}&=35\text{\textdegree}\mathrm{~C}\\ &=35+273\\ &=308\mathrm{~K}\\ D_{1}&=0.07\mathrm{~m}\\ R_{1}&=\dfrac{D_{1}}{2}\\ &=\dfrac{0.07m}{2}\\ &=0.035\mathrm{~m}\\ D_{2}&=0.08\mathrm{~m}\\ R_{2}&=\dfrac{D_{2}}{2}\\ &=\dfrac{0.08m}{2}\\ &=0.04\mathrm{~m}\\ \epsilon_{1} &=\epsilon_{2} =0.25\\ L &=0.3\mathrm{~m}\\ \end{align*}

Required:\textbf{Required:}

  • $\text{Heat loss from the coffee (qnet)(q_{net})}$

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