The radiosity of this surface is 299.95W/m².
<h3>
Calculation :</h3>
Given, € 0.8
Ta=0°C 273K
For opaque body., T = 0
a+ p = 1
€ = a (Kirchoff's Law)
p = (1-E)
Radiosity = leaving energy from the surface
= px 240 +€σT4
= (1-E) × 240 +0.8 x 5.67 x 10-8 (2734)
= (1-0.8) × 240 +0.8 x 5.67 × 10-8 (2734)
= 299.95W/m²
So, now after doing the calculations we can say that the radiosity of the surface was 299.95W/m².
To know more about Radiosity please click here : brainly.com/question/14937321
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Energy from food is measured in calories. We convert the calories into chemical energy.
Therefore, the answer is the 3rd choice.
Using pv=nRT
T= 22+273 K
P=1.15 Pa
R=8.31
V =5.5... (u did not write the unit so...)
Therefore n,mole= (1.15×5.5) ÷ (273+22)(8.31)
2. Left to right (smallest to largest) F-, Mg2+, Na+, and K+
5. Cs, Ga, As, N, O
6. Cs or H (probably cesium)
7. In or In 2+ (not sure which)
Hope this helps, I've never really seen questions like this before.
