A cubical block of mass 1.0 kg and edge 5.0 cm is heated to 227^0 C. It is kept in an evacuated chamber maintained at 27^0 C. Assuming that the block emits radiation like a blackbody, find the rate at which the temperature of the block will decrease. Specific heat capacity of the material of the block is 400 J $k g^{- 1} K^{- 1}$ .

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Solution

Since the cube can be assumend as black body

e = 1, $σ = 6.0 \times 10^{- 8} W / m^{2} - K^{4}$

A = $6 \times 25 \times 10^{- 4} m^{2}$

m = 1 kg, S = 400 J/kg - $^{\circ} K$

$T_{1} = 227^{\circ} C$ = 500 K

$T_{2} = 27^{\circ} C$ = 300 K.

ms= $\frac{d θ}{d t} = A σ e (T_{1}^{4} - T_{2}^{4})$

$\frac{d θ}{d t}$ =

$\frac{1 \times 6 \times 10^{- 8} \times 6 \times 25 \times 10^{- 4} [(500)^{4} - (300)^{4}]}{1 \times 400}$

= $\frac{36 \times 25 \times 544}{400} \times 10^{- 4}$

= 1224 $\times 10^{- 4}$

= ${0.12240}^{\circ} C / s$

= ${0.120}^{\circ} C / s$

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