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:
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.
