Question

The outer surface of a skier’s clothes of emissivity 0.7000.700 is at a temperature of 5.505.50 °C. Find the rate of radiation if the skier has a surface area of 1.601.60 m2 and the surroundings are at −20.0−20.0 °C.

Answer 1

Answer:

121.0 W

Explanation:

We use the equation for rate of heat transfer during radiation.

Q/t = σεA(T₂⁴ - T₁⁴)

Since temperature of surroundings = T₁ = -20.0°C = 273 +(-20) = 253 K, and temperature of skier's clothes = T₂ = 5.50°C = 273 + 5.50 = 278.5 K.

Surface area of skier , A = 1.60 m², emissivity of skier's clothes,  ε = 0.70 and σ =  5.67 × 10⁻⁸ W/m²K⁴ .

Therefore, the rate of heat transfer by radiation Q/t is

Q/t = σεA(T₂⁴ - T₁⁴) = (5.67 × 10⁻⁸ W/m²K⁴ ) × 0.70 × 1.60 m² × (278.5⁴ - 253⁴) = 6.3054 × (1918750544.0625) × 10⁻⁸ W = 1.2098 × 10² W = 120.98 W ≅ 121.0 W

Answer 2

Answer:

122.1 W.

Explanation:

Thermal radiation is defined as the electromagnetic radiation generated by the thermal motion of particles in matter.

Mathematically,

q = σ * ε * A * (Ts^4 - Ta^4)

where,

q = heat transfer per unit time (W)

σ = The Stefan-Boltzmann Constant = 5.6703 x 10^-8 (W/m^2.K^4)

Ts = absolute temperature of the surroundings in kelvins (K)

ε = emissivity of skier's clothes = 0.70

Ta = absolute temperature of the object in kelvins (K)

A = area of the emitting body (m2)

= 5.6703 x 10^-8 * 0.7 * 1.6 * (278.65^4 - 253.15^4)

= 122.1 W.