Question

A surface of area $1m^2$ kept perpendicular to the sun rays absorbs $1.8MJ$ of solar energy in one hour.
(a) What is the amount of electrical energy produced when a solar panel of area $5m^2$ is exposed perpendicular to the sun rays for $5h$. Take the efficiency of solar panel as 30%.
(b) Calculate the power generated in $kW$.

• A. $13.5MJ;0.75kW$
• B. $1.5MJ;0.75kW$
• C. $13.5MJ;1.75kW$
• D. $1.5MJ;1.75kW$

Answer 1

The correct option is A $13.5MJ;0.75kW$

The amount of energy absorbed by $1m^2$ area $=1.8MJ{h}^{-1}=1.8\times {10}^{6}J{h}^{-1}$
The area of solar panel $=5m^2$.
$\therefore$ The total amount of energy absorbed by solar panel in 1 hour $=5\times 1.8\times {10}^{6}J{h}^{-1}=9\times {10}^{6}J{h}^{-1}$
$\therefore$ The amount of energy absorbed in $5h=5\times 9\times {10}^{6}J=45\times {10}^{6}J$
$\therefore$ Efficiency of conversion $=30$%.
$\therefore$ Amount of electric energy produced $=30$% $[45\times {10}^6]J$
$=\frac{30}{100}\times [45\times {10}^6]J=1.5\times 9\times {10}^{6}J$
$=13.5\times {10}^{6}J=13.5MJ$.
$\therefore$ Power generated $\left(p\right)=\frac{E}{t}=\frac{energy}{unittime}=\frac{13.5\times {10}^{6}J}{5hour}$
$=\frac{13.5\times {10}^{6}J}{5\times 60\times 60s}=\frac{13.5}{36\times 5}\times {10}^{4}J{s}^{-1}$
$=\frac{13.5}{180}\times {10}^{4}W=\frac{13.5}{18}\times {10}^{3}W=0.75kW$
$\therefore$ Power generated $=0.75kW$.