# Step-by-Step Solution: Handle as a Fin Let's solve the problem step-by-step, using the assumptions and the "fin" approach suggested: --- ## **Assumptions** 1. **Steady-State Conditions:** Temperature does not change with time. 2. **1D Heat Transfer:** Heat transfer is along the length of the handle only (axial direction). 3. **Negligible Radiation:** Heat loss by radiation is much smaller than by convection. 4. **Adiabatic Tip:** The center (midpoint) of the handle is assumed insulated (no heat loss from the tip). 5. **Constant Properties:** \( k \) (thermal conductivity), \( h \) (convective coefficient) are constant. 6. **Symmetry:** Only half the handle is analyzed as one fin of length 8 inches. --- ## **Step 1: Gather Known Values** - **Length of one fin:** \( L = 8 \) inches = .2032 m - **Diameter of rod:** \( d = .25 \) in = .00635 m - **Ambient temperature:** \( T_\infty = 24^\circ C \) - **Base temperature:** \( T_b = 160^\circ C \) - **Thermal conductivity:** \( k = 52 \) W/m·°C - **Convective coefficient:** \( h = 12 \) W/m²·°C --- ## **Step 2: Fin Equation and Parameters** ### **Area and Perimeter** - **Cross-sectional area, \( A_c \):** \[ A_c = \frac{\pi d^2}{4} = \frac{\pi (.00635)^2}{4} = 3.17 \times 10^{-5} \text{ m}^2 \] - **Perimeter, \( P \):** \[ P = \pi d = \pi \times .00635 = .01995 \text{ m} \] ### **Fin Parameter, \( m \):** \[ m = \sqrt{\frac{hP}{kA_c}} \] Plug in the values: \[ m = \sqrt{\frac{12 \times .01995}{52 \times 3.17 \times 10^{-5}}} \] \[ m = \sqrt{\frac{.2394}{.0016484}} \] \[ m = \sqrt{145.28} \] \[ m \approx 12.05 \ \text{m}^{-1} \] --- ## **Step 3: Fin Equation for Insulated Tip** Temperature at distance \( x \) from the base (door): \[ \frac{T(x) - T_\infty}{T_b - T_\infty} = \frac{\cosh\left[m(L-x)\right]}{\cosh(mL)} \] - At **fin tip** (\( x = L \)), \( T(L) \) is the coolest temperature at the handle's center. --- ## **Step 4: Calculate Coolest Temperature \( T_{tip} \)** Plug \( x = L \): \[ \frac{T_{tip} - T_\infty}{T_b - T_\infty} = \frac{\cosh[]}{\cosh(mL)} = \frac{1}{\cosh(mL)} \] Calculate \( mL \): \[ mL = 12.05 \times .2032 = 2.45 \] \[ \cosh(2.45) = \frac{e^{2.45} + e^{-2.45}}{2} \approx \frac{11.58 + .0865}{2} = 5.83 \] So, \[ \frac{T_{tip} - 24}{160 - 24} = \frac{1}{5.83} \] \[ T_{tip} - 24 = \frac{136}{5.83} \] \[ T_{tip} - 24 = 23.32 \] \[ T_{tip} = 24 + 23.32 = 47.3^\circ C \] --- ## **Step 5: How Close Can You Grasp Without Being Burned?** Burns occur at \( 65^\circ C \). Let's solve for \( x \) where \( T(x) = 65^\circ C \): \[ \frac{65 - 24}{160 - 24} = \frac{\cosh[m(L-x)]}{\cosh(mL)} \] \[ \frac{41}{136} = .3015 \] \[ \cosh[m(L-x)] = .3015 \times 5.83 = 1.756 \] \[ m(L-x) = \cosh^{-1}(1.756) = \ln(1.756 + \sqrt{1.756^2 - 1}) \approx \ln(1.756 + 1.461) = \ln(3.217) = 1.169 \] \[ L-x = \frac{1.169}{m} = \frac{1.169}{12.05} = .097 \text{ m} \] \[ x = L - .097 = .2032 - .097 = .106 \text{ m} = 10.6 \text{ cm} = 4.2 \text{ in} \] --- ## **Step 6: Suggestions for Improving Design** - **Increase the length of handle (L):** Increases the distance from the hot door. - **Increase diameter:** Increases cross-sectional area, lowering temperature. - **Use a material with lower thermal conductivity (k):** Reduces heat transfer. - **Add insulation or a plastic/rubber grip at the handle's center.** --- ## **Final Answers** ### **1. Coolest temperature of the handle (center):** \[ \boxed{47.3^\circ C} \] ### **2. How close to the door can you grasp without being burned (at 65°C)?** \[ \boxed{4.2 \text{ inches from the door}} \] ### **3. Design improvements:** - Increase handle length - Increase diameter - Use materials with lower thermal conductivity - Add insulation or grip at the center