A potato that may be approximated as a \(5.7-\mathrm{cm}\) solid sphere with the properties \(\rho=910 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=4.25 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\), \(k=0.68 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and \(\alpha=1.76 \times 10^{-7} \mathrm{~m}^{2} / \mathrm{s}\). Twelve such potatoes initially at \(25^{\circ} \mathrm{C}\) are to be cooked by placing them in an oven maintained at \(250^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(95 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The amount of heat transfer to the potatoes by the time the center temperature reaches \(100^{\circ} \mathrm{C}\) is (a) \(56 \mathrm{~kJ}\) (b) \(666 \mathrm{~kJ}\) (c) \(838 \mathrm{~kJ}\) (d) \(940 \mathrm{~kJ}\) (e) \(1088 \mathrm{~kJ}\)

Understanding the lumped capacitance method is crucial when dealing with heat transfer problems that involve temperature changes in objects. This method simplifies the complex heat transfer calculations by assuming that the temperature within a solid object is uniform at any given instant in time. This uniformity implies no temperature gradient within the object, effectively treating it as a 'lump' of energy storage.

For the lumped capacitance method to be valid, the object must have a small Biot number (Bi), a dimensionless quantity given by \[\begin{equation}Bi = \frac{h L_c}{k}\end{equation}\] where \( h \) is the heat transfer coefficient, \( L_c \) is the characteristic length (often the volume to surface area ratio of the object), and \( k \) is the thermal conductivity. A Biot number less than 0.1 usually indicates the lumped capacitance method can be applied. In our potato problem, we assume that the Biot number is small enough for this to be the case, which simplifies the process of calculating the heat transfer.

Thermal conductivity (\( k \)) is a measure of a material's ability to conduct heat. It indicates how easily heat flows through a material due to a temperature gradient. In the context of the potato, its thermal conductivity value influences how quickly it can absorb heat from the oven's environment.

Highly conductive materials, such as metals, allow heat to distribute quickly, while low-conductivity materials, like the potatoes in our example, tend to heat up more slowly. The unit of measurement is watts per meter per Kelvin \( \left( \frac{W}{m \cdot K} \right) \). The value of thermal conductivity plays a pivotal role in the heat transfer rate and is an essential parameter in many heat transfer equations, including those used in the lumped capacitance method.

The heat transfer coefficient (\( h \)) represents the rate of heat transfer between a solid surface and a fluid per unit surface area per unit temperature difference. It is influenced by the nature of the fluid, the surface roughness, the flow velocity, and the type of heat transfer (convection or conduction). In our exercise, the value of \( h = 95 \frac{W}{m^2 \cdot K} \) implies the convective heat transfer rate from the oven to the potato.

A high heat transfer coefficient indicates efficient heat exchange, causing the potato to reach the desired temperature faster. Knowing this value is essential for engineers and designers when calculating heat transfer in various applications to ensure materials are heated or cooled effectively.