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 temperature $20^{o} C$ . The emissivity of the surface $= 0.80$ and $σ = 6.0 \times 10^{- 8} W / m^{2} - K^{4}$ .

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Solution

Given

Area, $A = 12 \times 10^{- 4}$ m²

Temperature, T = 273 + 20 = 293K

Emisivity, e = 0.80

Rate of emission of heat , R $= A e σ T ⁴$

$R = 12 \times 10^{- 4} \times 0.80 \times 6 \times 10^{- 8} \times (293) ⁴$

R = 0.42J

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Q. Find the heat radiated per second by a body of surface area

12 c m^{2}

kept in thermal equilibrium in a room at temperature

20^{o} C

. The emissivity of the surface is

0.8

and

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

Assume that the total surface area of a human body is 1.6 $m^{2}$ and that it radiates like an ideal radiator. Calculate the amount of energy radiated per second by the body if the body temperature is 37 $^{\circ}$ C.Stefan constant $σ$ is $6.0 \times 10^{- 8} W / m^{2} - K^{4}$ .

An electric heater emits 1000 W of thermal radiation. The coil has a surface area of 0.020 $m^{2}$ . Assuming that the coil radiates like a blackbody, find its temperature. $σ = 6.00 \times 10^{- 8} W / m^{2} - K^{4}$ .

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Calculate the amount of heat radiated per second by a body of surface area 12cm2 kept in thermal equilibrium in a room temperature 20oC. The emissivity of the surface =0.80 and σ=6.0×10−8W/m2−K4.

GivenArea, A=12×10−4 m²Temperature, T = 273 + 20 = 293KEmisivity, e = 0.80Rate of emission of heat , R =AeσT⁴ R=12×10−4×0.80×6×10−8×(293)⁴ R = 0.42J

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 temperature $20^{o} C$ . The emissivity of the surface $= 0.80$ and $σ = 6.0 \times 10^{- 8} W / m^{2} - K^{4}$ .

Given

Area, $A = 12 \times 10^{- 4}$ m²

Temperature, T = 273 + 20 = 293K

Emisivity, e = 0.80

Rate of emission of heat , R $= A e σ T ⁴$

$R = 12 \times 10^{- 4} \times 0.80 \times 6 \times 10^{- 8} \times (293) ⁴$

R = 0.42J