Calculate the solar energy incident on a reflecting surface in one second

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SUMMARY

The discussion focuses on calculating the solar energy incident on a reflecting surface for a solar cooker with a surface area of 0.75 m² and an incident solar power of 800 W/m². The user correctly identifies that the energy arriving at the curved surface each second is 600 Joules, calculated by multiplying the power per unit area by the area of the surface. For the total energy reflected towards the pot in one hour, the user concludes that it involves finding 76% of the total energy calculated over 3600 seconds, resulting in an effective energy output of 1,296,000 Joules.

PREREQUISITES
  • Understanding of solar power metrics, specifically W/m²
  • Basic knowledge of energy calculations involving Joules and Watts
  • Familiarity with the concept of percentage for energy reflection
  • Ability to perform unit conversions and time calculations
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  • Study the principles of solar energy conversion and efficiency
  • Learn about the physics of energy reflection and absorption
  • Explore calculations involving energy over time, specifically in Joules
  • Investigate the design and optimization of solar cookers
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Students preparing for exams in physics, renewable energy enthusiasts, and individuals interested in solar cooker design and efficiency calculations.

Richie Smash
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Homework Statement


First I will say, I know I post here a lot, but that is because I have exams coming up soon and I am a beginner, and It's crunch time, I'm self taught with only the help of friends and the internet. I'm currently doing a past examination as practice, but this one is much harder than the ones I've done before, and from every single post I've placed here, I have learned, I like how the helpers only give hints.

Ok without any further ado here is the question.

''A solar cooker focuses sunlight onto a pot containing food. The area of the reflecting surface is 0.75m2 and the solar power incident on it is 800Wm-2. Only 76% of the energy incident on the curved surface is reflected towards the pot.

Calculate:(i) The energy arriving at the curved surface each second.
(ii) the total energy reflected towards the pot in one hour

Homework Equations


KE=1/2*m*v2
W=J/S

The Attempt at a Solution


I would try to solve it but the problem is I don't have a time...
I'm thinking that ok the solar radiation is coming to the reflecting surface at 3.0*108m/s but... That's all I have, I'm unfamiliar with this Watt per metre squared.
 
Last edited by a moderator:
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A Watt is a Joule/second. 1 Watt/m2 means that if you take a square that is 1 m on the side, then every second that goes by, one Joule worth of energy is incident on the surface. Here you are given the surface. In (i) you are asked to find how much energy hits the surface in one second. Do you think you can answer that?
 
Ok I'm following a bit, that means in this question in one second 800 joules will hit a square 1m on the side.

But the area here is 0.75m2
So what I have to do is, 800 *0.75 and I will get 600Joules?

And now for one hour I would do 600*3600 and just simply find 76% of that and that is the answer.
 
Last edited:
Richie Smash said:
600*60
How many seconds in an hour?

As an aside, the question says the curved surface has an area of .75m2. Presumably it means that is the cross-sectional area, i.e. the area of a flat surface with the same perimeter. Otherwise there is not enough information. We would need, e.g., the distance from mirror to pot.
 
Ah yes my mistake 600*3600
 

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