Question

a metallic sphere having radius 0.08 m and mass m = 10kg is heated to a temperature of 227°c and suspended inside a box whose walls are at a temperature of 27°c. the maximum rate at which it's temperature will fall is? take e=1, stefan's constant=5.8×10^-8 and specific heat of the metal s=90, J=4.2 joules/calorie​

Answer 1

using formula,

$\quad ms\left(-\frac{dT}{dt}\right)=\sigma eA(T^4-T_s^4)$

where m is mass of metallic sphere, e.g.,m = 10kg,

s is specific heat of the metal , s = 90 × 4.2 J/°C.Kg

-dT/dt is the rate of falling in temperature.

$\sigma$ is Stefan's constant e.g., $\sigma$= 5.8 × 10^-8 J/K⁴m²s

e is emissivity . i.e., e = 1

A is cross sectional area, A = π(0.08)² m²

and T = 227°C =227 + 273 = 500K

Ts = 27°C = 27 + 273 = 300K

now, 10 × 4.2 × 90 × (-dT/dt) = 5.8 × 10^-8 × 1 × π(0.08)² × (500⁴ - 300⁴)

or, -dT/dt = {5.8 × 10^-8 × π × 64 × 10^-4 × 16 × 10⁴ }/(42 × 90}

= 0.06°C/s

hence, falling in temperature at the rate of 0.06°C/s.