General ScienceThe surface area of a concentrator type solar heater is 10 m2. It can reflect 75%of the solar radiation incident on it while it absorbs the rest. Calculate the energy concentrated by heater at its focus in 1 hour. If the solar energy is delivered to it at rate of 0.38 kWm2.

The surface area of a concentrator type solar heater is $10 m^{2}$ . It can reflect $75 %$ of the solar radiation incident on it while it absorbs the rest. Calculate the energy concentrated by heater at its focus in $1 hour$ . If the solar energy is delivered to it at rate of $0.38 \frac{kW}{m^{2}}$ .

A
$10.26 \times 10^{6} Joules$
B
$1.26 \times 10^{6} Joules$
C
$88.26 \times 10^{6} Joules$
D
$19.26 \times 10^{6} Joules$

Solution:

The energy concentrated by heater at its focus in

1 hour

is given by

10.26 \times 10^{6} Joules

Convert

\frac{kW}{m^{2}} to \frac{joule}{\sec}

1 \frac{joule}{\sec} = 1 watt

0.38 \frac{kW}{m^{2}} = 0.38 \times 10^{3} \frac{\frac{joule}{\sec}}{m^{2}}

Energy concentrated = power \times area \times radiation incident \times time in sec

= 0.38 \times 10^{3} \times 10 \times 75 \times 3600

= 10.26 \times 10^{6} Joules