Question

A certain solar panel is capable of absorbing 750J of light energy every second and converting 150J of the energy into electrical energy.
(a): How much energy is wasted in the form of heat by the solar panel every second.
(b): What is the efficiency of this solar panel?

Answer 1

Given :

• Light energy absorbed per second by solar panel = 750J
• Electrical energy produced per second by solar panel = 150J

To find :

(a) Energy is wasted in the form of heat by the solar panel every second.

(b): Efficiency of this solar panel.

Solution :

As we can see in question that there is a solar panel which absorb light energy and convert/produce electrical energy from it and some amount of absorbed light energy is wasted as heat

Therefore,

➝ Light energy absorbed per second = Electrical energy produced per second + Heat lost per second

➝ Heat lost per second = Light energy absorbed per second - Electrical energy produced per second

➝ Heat lost per second = 750J - 150J

➝ Heat lost per second = 600J

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$\sf Efficency \: of \: solar \: panel \: = \: \dfrac{ Electrical \: energy \: produced \: per \: second }{ \: Light \: energy \: absorbed \: per \: second} \: \times 100 \: \%$

$\sf \implies Efficency \: of \: solar \: panel \: = \: \dfrac{150}{750} \times 100 \: \%$

$\sf \implies Efficency \: of \: solar \: panel \: = \: \dfrac{15000}{750} \%$

$\sf \implies Efficency \: of \: solar \: panel \: = \: 20 \%$

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ANSWER :

(a) 600 J

(b) 20%