The surface area of a concentrator type solar heater is $5 m^{2}$ . It reflects $80 %$ of solar radiation. The energy concentrated by the heater in $2$ hours is ____ $\times 1 0^{3}$ $kj$ . If solar energy delivered to it at a rate of $0.4 kj / m^{2} s$ .

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Detailed Solution

The surface area of a concentrator type solar heater is

5 m^{2}

. It reflects

80 %

of solar radiation. The energy concentrated by the heater in

2

hours is

2.88 \times 1 0^{3}

kj

. If solar energy delivered to it at a rate of

0.4 kj / m^{2} s

.
Since the concentrator reflects

80 %

of solar radiation, then it absorbs only

20 %

of the solar radiation.
Here,

A = 5 m^{2}, t = 2 \times 60 \times 60, k = 0.4 kj / m^{2} s

Then the energy concentrated by the heater is

= 5 \times 7200 \times 0.4 \times 0.2

= 2.88 \times 1 0^{3} kj

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