GATE 2026 ME – Question 35
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 convection heat transfer coefficient of 500 W/(m$^2$K). For the given range of temperatures, the thermophysical properties of the material of the sphere are: density $\rho = 3000$ kg/m$^3$, thermal conductivity $k = 10$ W/(mK), and specific heat $c = 1000$ J/(kgK). Neglecting radiation heat transfer, the time required for the centre of the sphere to cool from 400 °C to 50 °C is ________ seconds (*rounded off to 2 decimal places*).
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Correct answer: 25.26 to 25.52
Explanation
The characteristic length is $\frac{V}{A} = \frac{r}{3} = 0.001667$ m, so the Biot number is $\frac{500 \times 0.001667}{10} = 0.083$. This is below 0.1, so the lumped model applies. The time constant is $\tau = \frac{\rho c (V/A)}{h} = \frac{3000 \times 1000 \times 0.001667}{500} = 10$ s. Then $\frac{50 - 20}{400 - 20} = e^{-t/\tau}$, so $t = 10\ln\frac{380}{30} = 25.39$ s.