Calculate the amount of heat radiated per second by a body of surface area 12 $c m^{2}$ kept in thermal equilibrium in a room at temperature $20^{0}$ C. The emissivity of the surface = 0.80 and $σ = 6.0 \times 10^{- 8} W m^{- 2} K^{- 4}$

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Solution

A = $12 c m^{2} = 12 \times 10^{- 8} \times 10^{4} m^{2}$ ,

T = $20^{0} C = 293^{0}$ K, e =0.8

$σ = 6 \times 10^{- 8} W / m^{2} - K^{4}$

q = $A e S T^{4}$

= $12 \times 10^{- 4} \times 0.8 \times 6 \times 10^{- 8} \times (293)^{4}$

= 0.4245 = 0.42

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Calculate the amount of heat radiated per second by a body of surface area 12 cm2 kept in thermal equilibrium in a room at temperature 200 C. The emissivity of the surface = 0.80 and σ=6.0×10−8Wm−2K−4

A = 12cm2=12×10−8×104m2, T = 200C=2930 K, e =0.8 σ=6×10−8W/m2−K4 q = AeST4 = 12×10−4×0.8×6×10−8×(293)4 = 0.4245 = 0.42

Calculate the amount of heat radiated per second by a body of surface area 12 $c m^{2}$ kept in thermal equilibrium in a room at temperature $20^{0}$ C. The emissivity of the surface = 0.80 and $σ = 6.0 \times 10^{- 8} W m^{- 2} K^{- 4}$

A = $12 c m^{2} = 12 \times 10^{- 8} \times 10^{4} m^{2}$ ,

T = $20^{0} C = 293^{0}$ K, e =0.8

$σ = 6 \times 10^{- 8} W / m^{2} - K^{4}$

q = $A e S T^{4}$

= $12 \times 10^{- 4} \times 0.8 \times 6 \times 10^{- 8} \times (293)^{4}$

= 0.4245 = 0.42