Can I see a photo or can you comment the answer choices.
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
1. circle graph
2. Bar graph
3. line graph
hope this helps
I think the answer is A.........
