Two rectangular surfaces are perpendicular to each other with a common edge that measures 1.6 m. Surface one is 0.8 m wide, has an emissivity of 0.75, and is maintained at 450 K. Surface two is a blackbody that measures 1.2 m high and is maintained at 700 K. If the surrounding surfaces are taken to be at 290 K with an emissivity of 0.85, what is the net rate of radiative heat transfer between the two surfaces and between surface one and the surroundings

Answer8.0m;

Answerthe energy required is 1.60 × 10⁻¹⁸ j, to three significant figures.explanationwe define electric potential as the ratio electric potential energy to the charge,    then the following gives us the potential difference,      the formula  may be on your formula sheet.the change in electric potential energy would be the energy that would needed to move it from point a to point b. solving for that, we get    using q = 8.00  × 10⁻¹⁹ c and  δv = 2.00 v, we get    the energy required is 1.60 × 10⁻¹⁸ j, to three significant figures.

Cis the answer  hope this : d