Question

d. (4 Points)A photovoltaic cell converts light into electrical energy. Suppose a certain photovoltaic cell is only 43.5% efficient, in other words, that 43.5% of the light energy is ultimately recovered. If the energy output of this cell is used to heat water, how many 512 nm photons must be absorbed by the photovoltaic cell in order to heat 10.0 g of water from 20.0°C to 30.0°? 07 The specific heat of water is 4.184 J/g-"C.

Answer 1

energy associated with each 512nm photon

$E=\frac{hc}{\lambda }$

=

(6.626 * 10-34) * (3 * 108) (512 * 10-9)

=
3.8824*10-12
J

so the energy associated with n photons is

n* 3.8824 * 10-19

but its only 43.5% efficient

which gives the total energy obatained as

* 3.8824 + 10-19 +0.435

=(total energy obtained from one photon)

1.69 * 10-13

The total energy required to heat water from 20 to 30 is

= MCA

=10*4.184*(30-20)

=418.4J

by equating the two we get

m = 418.4 1.69 * 10-19

     =
2.4757*1021
photons