Question

A surface area $1\,{{\text{m}}^2}$ kept perpendicular to the sun rays absorbs $1.8\,{\text{MJ}}$ of solar energy in one hour.
(a) What is the amount of electrical energy produced when the solar panel of area $5\,{{\text{m}}^2}$ is exposed perpendicular to the sun rays for $5\,{\text{h}}$. Take the efficiency of the solar panel as 30%.
(b) Calculate the power generated in ${\text{kW}}$.
A. $13.5\,{\text{MJ; 0}}{\text{.75 kW}}$
B. $1.5\,{\text{MJ; 0}}{\text{.75 kW}}$
C. $13.5\,{\text{MJ; 1}}{\text{.75 kW}}$
D. $1.5\,{\text{MJ; 1}}{\text{.75 kW}}$

Answer 1

First calculate the amount of solar energy absorbed by the solar panel in 5 hour by the solar panel of the given surface area. Then using the efficiency of the solar panel, calculate the electric energy produced by the solar panel. Now use the formula for the power in terms of the energy and time and calculate the power generated in kilowatt.

Formulae used:
The power $P$ is given by
$P = \dfrac{Q}{t}$ …… (1)
Here, $Q$ is the heat energy and $t$ is the time.

Complete step by step answer:
We have given that the solar energy absorbed by the solar panels of surface area $1\,{{\text{m}}^2}$ in one hour is $1.8\,{\text{MJ}}$.

(a) We have asked to calculate the amount of electrical energy produced by the solar panel.The amount of solar energy absorbed by the solar panels in 5 hours is
$Q = 5 \times 1.8\,{\text{MJ}}$
$\Rightarrow Q = 9\,{\text{MJ}}$
The amount of solar energy absorbed by the solar panel of surface area is
$\Rightarrow Q = \left( {5\,{{\text{m}}^2}} \right)\left( {9\,{\text{MJ}}} \right)$
$\Rightarrow Q = 45\,{\text{MJ}}$
$\Rightarrow Q = \left( {45\,{\text{MJ}}} \right)\left( {\dfrac{{{{10}^6}\,{\text{J}}}}{{1\,{\text{MJ}}}}} \right)$
$\Rightarrow Q = 45 \times {10^6}\,{\text{J}}$
Hence, the total amount of solar energy absorbed by the solar panels is $45 \times {10^6}\,{\text{J}}$.

We have given that the efficiency of the solar panel is 30%. This shows that only 30% of the solar energy absorbed by the solar panel is converted into electrical energy.Hence, the amount of the electrical energy produced by the solar panel is
$E = \left( {45 \times {{10}^6}\,{\text{J}}} \right) \times \dfrac{{30}}{{100}}$
$\Rightarrow E = 13.5 \times {10^6}\,{\text{J}}$
$\Rightarrow E = \left( {13.5 \times {{10}^6}\,{\text{J}}} \right)\left( {\dfrac{{{{10}^{ - 6}}\,{\text{MJ}}}}{{1\,{\text{J}}}}} \right)$
$\therefore E = 13.5\,{\text{MJ}}$

Hence, the electrical energy produced by the solar panel is $13.5\,{\text{MJ}}$.

(b) Let us now calculate the power generated by the solar panels.Rewrite equation (1) for the power.
$P = \dfrac{E}{t}$
Substitute $13.5\,{\text{MJ}}$ for $E$ and $5\,{\text{h}}$ for $t$ in the above equation.
$P = \dfrac{{13.5\,{\text{MJ}}}}{{5\,{\text{h}}}}$
$\Rightarrow P = \dfrac{{13.5\,{\text{MJ}}}}{{\left( {5\,{\text{h}}} \right)\left( {\dfrac{{3600\,{\text{s}}}}{{1\,{\text{h}}}}} \right)}}$
$\Rightarrow P = 0.00075\,{\text{MW}}$
$\Rightarrow P = \left( {0.00075\,{\text{MW}}} \right)\left( {\dfrac{{{{10}^3}\,{\text{kW}}}}{{1\,{\text{MW}}}}} \right)$
$\therefore P = 0.75\,{\text{kW}}$

Hence, the power generated is $0.75\,{\text{kW}}$.

Note: The students should be careful while calculating the power generated by the solar panels. The time for which the power generated is to be calculated is in the unit hour. So the unit of this time should be converted to the SI system of units that is from hour to seconds. Then only one can convert the obtained power in the unit kilowatt.