A sphere of radius 5 mm is initially in equilibrium at 400°C in a furnace. It is suddenly removed from the furnace and dipped in a well-stirred water bath at 20°C, with a convective heat transfer coefficient of 500 W/m²K. For the given range of temperatures, the thermophysical property of the material of the sphere are density = 3000 kg/m³, k = 10 W/mK, c = 1000 J/kgK. Neglecting radiation heat transfer, the time required for the center of the sphere to cool from 400°C to 50°C is ____ sec. (Round off to two decimal places)

The correct answer is 25.39.

To determine the time required for the center of the sphere to cool from 400°C to 50°C, we first need to check if the lumped parameter analysis (lumped capacitance method) is applicable. This is determined by calculating the Biot number (Bi).

The Biot number for a sphere is defined as:

$Bi=\frac{h{L}_c}{k}$

where:
- $h$ is the convective heat transfer coefficient ($500{\text{W/m}}^2\text{K}$)
- $k$ is the thermal conductivity of the material ($10\text{W/mK}$)
- $L_c$ is the characteristic length of the sphere

For a solid sphere, the characteristic length $L_c$ is the ratio of its volume to its surface area:

$L_c=\frac{V}{A_s}=\frac{\frac{4}{3}\pi R^3}{4\pi R^2}=\frac{R}{3}$

Given the radius $R=5\text{mm}=5\times {10}^{-<!--</mo--><mn>3</mn>}\text{m}$, we have:

$L_c=\frac{5\times {10}^{-3}}{3}\text{m}$

Substituting the values to find the Biot number:

$Bi=\frac{500\times \left(\frac{5\times {10}^{-3}}{3}\right)}{10}=\frac{2.5/3}{10}=\frac{2.5}{30}\approx 0.0833$

Since the Biot number $Bi<0.1$, the temperature distribution within the sphere is sufficiently uniform, and we can apply the lumped parameter analysis to find the temperature as a function of time.

The temperature variation with time $t$ in a lumped system is given by the relation:

$\frac{T\left(t\right)-T_{\infty }}{T_i-T_{\infty }}=e^{-\left(\frac{h{A}_s}{\rho Vc}\right)t}$

where:
- $T\left(t\right)=50°\text{C}$ (final temperature)
- $T_{\infty }=20°\text{C}$ (water bath temperature)
- $T_i=400°\text{C}$ (initial temperature)
- $\rho =3000{\text{kg/m}}^3$ (density)
- $c=1000\text{J/kgK}$ (specific heat capacity)

Using $\frac{A_s}{V}=\frac{3}{R}$, the exponent coefficient simplifies to:

$\frac{h{A}_s}{\rho Vc}=\frac{3h}{\rho Rc}$

Substituting the values:

$\frac{3h}{\rho Rc}=\frac{3\times 500}{3000\times \left(5\times {10}^{-3}\right)\times 1000}=\frac{1500}{15000}=0.1{\text{s}}^{-1}$

Now, substitute this back into the temperature equation:

$\frac{50-20}{400-20}=e^{-0.1t}$

$\frac{30}{380}=e^{-0.1t}$

Taking the natural logarithm ($ln$) of both sides:

$ln\left(\frac{3}{38}\right)=-0.1t$

$-2.53897=-0.1t$

$t=\frac{2.53897}{0.1}\approx 25.39\text{seconds}$

Thus, the time required for the center of the sphere to cool from 400°C to 50°C is approximately 25.39 seconds.