Answer:
1247.24J or 1.25kJ
Step-by-step explanation:
The rate of heat transfer by emitted radiation is determined by the Stefan-Boltzmann law of radiation:
Qnet/t =&eA(Ts⁴-Ta⁴)
Qnet/t is the net energy transfer per time
& is the Stefan-Boltzmann constant 5.671 × 10^-8 kg s^-3 K-4 (try to use correct sign)
e is the emissivity of the object i.e is a measure of how well it radiates. In this case, 0.9
A is the area of the radiating surface; 1.5m²
Ts is the absolute skin temperature: 35+273=308K
Ta is the absolute ambient (bedroom) temperature: 20+273=293K
Qnet/t= 5.671×10^-8 ×0.9×1.5×(308⁴-293⁴)
Qnet/t=124.724J/s
For 10 mins or 600sec
Qnet=124.724×t
Qnet= 1247.24J or 1.25kJ