Consider a stainless steel spoon \(\left(k=8.7 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{F}\right)\) partially immersed in boiling water at \(200^{\circ} \mathrm{F}\) in a kitchen at \(75^{\circ} \mathrm{F}\). The handle of the spoon has a cross section of \(0.08\) in \(\times\) \(0.5\) in, and extends 7 in in the air from the free surface of the water. If the heat transfer coefficient at the exposed surfaces of the spoon handle is \(3 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2} \cdot{ }^{\circ} \mathrm{F}\), determine the temperature difference across the exposed surface of the spoon handle. State your assumptions. Answer: \(124.6^{\circ} \mathrm{F}\)

The problem essentially involves finding the temperature difference across the exposed surface of the spoon handle due to heat transfer through conduction and convection.

The following information is given in the problem: - Thermal conductivity of the stainless steel (k): 8.7 Btu/h·ft·°F - Boiling water temperature (T_boil): 200°F - Kitchen temperature (T_kitchen): 75°F - Cross-sectional dimensions of the handle: 0.08 in × 0.5 in - Length of the handle exposed in the air: 7 in - Heat transfer coefficient of the exposed surfaces of the handle (h): 3 Btu/h·ft²·°F Before we start solving the problem, we need to convert the dimensions of the handle to feet.

To convert the dimensions of the handle to feet, divide the given dimensions by 12 (1 foot = 12 inches). Cross-sectional area (A): \((0.08 \times 0.5)/144 = 2.7778 \times 10^{-4} \,\text{ft}^2\) Length of the exposed handle (L): \(7/12 = 0.5833 \,\text{ft}\)

To find the heat transfer rate (Q_dot) through conduction in the handle, use the following equation: $$ Q_\text{dot} = k \cdot \frac{A\cdot \Delta \text{T}}{L} $$ However, we first need to find the temperature difference (ΔT) across the handle.

To find the temperature difference (ΔT) across the handle, we first need to find the conducted heat transfer rate. We know that the heat transfer rate is the same through conduction and convection. $$ Q_\text{dot} = h \cdot A \cdot\Delta \text{T}_\text{conv} $$ From step 4, we know: $$ Q_\text{dot} = k \cdot \frac{A\cdot \Delta \text{T}_\text{cond}}{L} $$ Equating the two expressions for Q_dot and solving for the temperature difference across the exposed surface of the spoon handle: $$ \Delta \text{T}_\text{conv} = \frac{k\cdot L}{h}\cdot \Delta \text{T}_\text{cond} $$ In order to find the temperature difference across the exposed surface of the spoon handle, we first need to find \(\Delta \text{T}_\text{cond}\): $$ \Delta \text{T}_\text{cond} = \frac{Q_\text{dot}\cdot L}{k\cdot A} $$ $$ \Delta \text{T}_\text{cond} = \frac{T_\text{boil} - T_\text{kitchen}}{\frac{k\cdot L}{h\cdot A} + 1} $$ Now plug in the given values and calculate: $$ \Delta \text{T}_\text{cond} = \frac{200 - 75}{\frac{8.7\cdot 0.5833}{3\cdot2.7778 \times 10^{-4}} + 1} $$ With this, we find the temperature difference across the stainless steel handle due to conduction: $$ \Delta \text{T}_\text{cond} = 69.4\,^{\circ}\text{F} $$ Now, we will find the temperature difference at the exposed surface using the above derived equation: $$ \Delta \text{T}_\text{conv} = \frac{8.7\cdot 0.5833}{3}\cdot 69.4 $$ $$ \Delta \text{T}_\text{conv} = 124.6\,^{\circ}\text{F} $$ The temperature difference across the exposed surface of the spoon handle is 124.6°F.