Question

8. A photovoltaic cell converts light in 73.6% efficient, in other words, that energy output of this cell is used to he the photovoltaic cell in order to heat 4500. CHE 111 light into electrical energy. Suppose a certain photovoltaic cell is only vords, that 73.6% of the light energy is ultimately recovered. If the is used to heat water, how many 633nm photons must be absorbed by in order to heat 4500.0 g of water from 0.0°C to 214°C?

Answer 1

Numbdr of photons that must be absorbed bybthe photovoltaic cell is 1.744×10²⁵

Energy of the photon, Epheton = ho h=6626x10 34 Is =hl = 6.626x1034J8x 2.998 x10 vas c= speed of light = 2.998 x 108ms'? - 63x109 1 hp = 67 = 3.138 x 10'9 J x= wavelength wa frequency Enagy required to hear the water, E4,0 = mx crot m= mass C=4.184 /gec = 4500.0g * 41843 x(214-0.05% AT = Teimperatine l m creased 9029192 J Number of photons needed for 4029192 ] - EH2O = 4029192 J Ephoten 3.138x1095 = 1.284 x 10 25

Effeciency I cell = 7316% This means, when the cell absorb loo phetons only 73.6 photons are able to used for work. So, the actual number of photons that the Cell needed to absorb - 25 0 x 1.284x10 photons 73.6 25 = 11.744 x10% photons