Question

Using a cylinder, a sphere, and a cube as examples, show that the rate of heat transfer is inversely proportional to the nominal size of the object. That is, heat transfer per unit area increases as the size of the object decreases.

Short answer

Answer: The rate of heat transfer per unit area (represented by the ratio of surface area to volume) is inversely proportional to the size (diameter) of the object for all three shapes (cylinder, sphere, and cube). As the size of the object decreases, the heat transfer per unit area increases, and vice versa.

Step-by-step solution

Derive Surface Area and Volume Formulas for Each Shape

For each of the three shapes - the cylinder, the sphere, and the cube - we must derive two equations: one for the surface area and one for the volume. Let's denote the nominal size (diameter) of each shape as 'D'. Cylinder: Surface Area (A_cylinder) = \(2\pi (D/2)^2 + 2\pi (D/2)h\) Volume (V_cylinder) = \(\pi (D/2)^2 h\) Sphere: Surface Area (A_sphere) = \(4\pi (D/2)^2\) Volume (V_sphere) = \(\frac{4}{3}\pi (D/2)^3\) Cube: Surface Area (A_cube) = \(6D^2\) Volume (V_cube) = \(D^3\)

Calculate Ratio of Surface Area to Volume for Each Shape

To determine the heat transfer per unit area, we must find the ratio of surface area to volume for each shape. Let's calculate the ratios using the derived equations from Step 1. Cylinder: (A_cylinder/V_cylinder) = \(\frac{2\pi (D/2)^2 + 2\pi (D/2)h}{\pi (D/2)^2 h}\) Sphere: (A_sphere/V_sphere) = \(\frac{4\pi (D/2)^2}{\frac{4}{3}\pi (D/2)^3}\) Cube: (A_cube/V_cube) = \(\frac{6D^2}{D^3}\)

Simplify Ratio Expressions and Compare for Different Sizes

Now, let's simplify the expressions for the ratios of surface area to volume that were calculated in Step 2. After simplification, we will compare the expressions for different sizes (diameters). Cylinder: (A_cylinder/V_cylinder) = \(\frac{2\pi (D/2)^2 + 2\pi (D/2)h}{\pi (D/2)^2 h}\) = \(\frac{2}{D} + \frac{1}{h}\) Sphere: (A_sphere/V_sphere) = \(\frac{4\pi (D/2)^2}{\frac{4}{3}\pi (D/2)^3}\) = \(\frac{3}{D}\) Cube: (A_cube/V_cube) = \(\frac{6D^2}{D^3}\) = \(\frac{6}{D}\) Each expression for the ratio of surface area to volume depends on 'D', the nominal size (diameter) of the object. As 'D' decreases, the heat transfer per unit area, represented by the ratios of surface area to volume, will increase, as shown by the expressions above. This demonstrates that the rate of heat transfer is indeed inversely proportional to the size of the object, as required by the given exercise.