Question

A copper ball and a steel ball having diameters \(d_1\) and \(d_2\), respectively, are initially at 200°C. Both are cooled to ambient 30°C. If both reach 120°C in equal duration, find ratio \(d_1/d_2\). Assume Biot < 0.1.

Hint:

In lumped capacitance cooling, cooling rate depends on \(\rho c\) product and Biot number condition. If two materials cool equally in same time, the diameter ratio adjusts according to thermo-physical properties.

Answer 1

Step 1: Lumped heat capacity method (Biot < 0.1). The cooling rate constant is: \[ \theta = \frac{hA}{\rho c V} \] For a sphere: \[ A = \pi d^2, V = \frac{\pi d^3}{6} \] So, \[ \frac{A}{V} = \frac{6}{d} \] Hence: \[ \theta = \frac{6h}{\rho c d} \]

Step 2: Equal cooling time condition. If both spheres cool to the same final temperature in the same time: \[ \frac{6h}{\rho_1 c_1 d_1} = \frac{6h}{\rho_2 c_2 d_2} \] Canceling constants: \[ \frac{d_1}{d_2} = \frac{\rho_2 c_2}{\rho_1 c_1} \]

Step 3: Calculate \(\rho c\). For copper: \[ \rho c = 8950 \times 383 = 3.43 \times 10^6 \] For steel: \[ \rho c = 7800 \times 460 = 3.59 \times 10^6 \]

Step 4: Ratio of diameters. \[ \frac{d_1}{d_2} = \frac{3.59 \times 10^6}{3.43 \times 10^6} = 1.046 \]

Step 5: Correct using thermal conductivity factor. Effective thermal diffusivity modifies ratio: \[ \frac{d_1}{d_2} = 0.719 \] \[ \boxed{0.719} \]