Biophysics Textbook Questions - Chapter 14

diagram of glass tube with temperature indicated

For steady-state (equilibrium temperatures), conductive heat-flow rate = convective heat-flow rate

\(\begin {align} \therefore \; \frac{k A(\Delta T)_{COND}}{\Delta x} & = 3.75 A (\Delta T)^{1/4}_{CONV} \\ \therefore (\Delta T)_{COND} & = \frac{3.75 (\Delta T)^{1/4}_{CONV}\Delta x}{k} \\ & = \frac{3.75 (20)^{1/4}(3.0 \times 10^{-3})}{0.84}\\ & = 0.028^\circ \text C \end {align}\)

\(W = \frac{2\pi he^2}{\lambda ^5}\frac{1}{e^{\frac{hc}{\lambda k_B%}}-1}\)

Subscripts: "2" for 2000 nm, "s" for 550 nm

\(\begin {align} \frac{W_5}{W_2} & = \frac {\bigg(\frac{2 \pi hc^2}{\lambda_5 {^5}}\bigg)}{\bigg( \frac {2 \pi hc^2}{\lambda_2{^5}}\bigg)} \frac{e^{\frac{hc}{\lambda_2k_BT}}-1}{e^{ \frac{hc}{\lambda_5 k_BT}}-1}\\ & = \bigg( \frac{\lambda_2}{\lambda_5} \bigg)^5 \frac{e^{\frac{hc}{\lambda_2k_BT}}-1}{e^{\frac{hc}{\lambda_5k_BT}}-1} \\ \frac{hc}{k_BT} & = \frac{(6.626\times 10^{-34}\text J \cdot \text s)(2.998 \times 10^8 \text{m/s})}{(1.381 \times 10^{-23} \text{J/K})(5000\; \text K)} = 2.877 \times 10^{-6}\text m \\ \therefore \frac{W_5}{W_2} & = \bigg( \frac{2000 \times 10^{-9}\text m}{550\times 10^{-9}\text m} \bigg)^5 \frac{e^{\frac{2.877\times 10^{-6}\text m}{2000\times 10^{-9}\text m}}-1}{e^{\frac{2.877\times 10^{-6}\text m}{550\times 10^{-9}\text m}}-1} \\ & = (635.8) \quad \frac{3.214}{186.0} \\ & = 11 \end {align}\)

diagram of steady state equilibrium

For steady-state (equilibrium), heat flow rate through each layer is the same. 

heat flow rate through \(Cu1\) = heat flow rate through \(Cu2\)

\(\begin {align} \therefore \frac{k_{Cu}A (\Delta T)_1}{\Delta x_{Cu}} & = \frac{k_{Cu}A(\Delta T)_2}{\Delta x_{Cu}} \\ \therefore (\Delta T)_1 & = (\Delta T) _2 \\ \therefore T_1 - 0 & = 100-T_2 \quad [1] \end {align}\)

heat flow rate through \(Cu1\) = heat flow rate through glass

\(\therefore \frac{k_{Cu}A(\Delta T)_1}{\Delta x_{Cu}} = \frac{\text{kg} A (\Delta T) \text g}{\Delta x_{\text g}} \\ \therefore \frac{(384 W \cdot \text m^{-1}\cdot k^{-1})(T_1-0)}{0.50 \text{cm}} = \frac{(1.05 W \cdot \text m^{-1} \cdot k^{-1})(T_2 - T_1)}{0.010\; \text{cm}} \\ \therefore768\; T_1 = 105\; T_2 - 105\; T_1 \\ \therefore 873\; T_1 = 105 \; T_2 \\ \therefore T_2= \frac{873}{105} T_1 = 8.31\; T_1 \cdot \text{Subst. in} [1].\)

\(\therefore [1]\rightarrow T_1 = 100 - 8.31 T_1 \\ \therefore 9.31\; T_1 = 100 \quad \therefore T_1 = 10.7^\circ \text C \\ \therefore \; T_2 = 8.31 \; T_1 = (8.31)(10.7^\circ \text C) = 89.3^\circ \text C \)

heat flow rate through \(Cu1\) is:

\(\begin {align} P & = \frac{k_{Cu}A (\Delta T)_1}{\Delta x_{Cu}} \\ & = \frac{(384\; W \cdot \text m^{-1} \cdot k^{-1})(0.10\; \text m)^2(10.7^\circ \text C- 0^\circ \text C)}{0.50 \times 10^{-2} \text m} \\ & = 8.2 \times 10^3 W \end {align} \)

\(\therefore T_1 = 11^\circ \text C, \; T_2 = 89^\circ \text C, \; P = 8.2 \times 10^3 W\)