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

AI Thread Summary
The discussion centers on calculating the solar energy incident on a reflecting surface for a solar cooker. The reflecting surface area is 0.75 m², with solar power incident at 800 W/m², and 76% of the energy is reflected towards the pot. The initial calculation involves determining the energy hitting the surface in one second, which is found to be 600 Joules. For the total energy reflected towards the pot in one hour, the calculation involves multiplying the energy per second by 3600 seconds and then applying the 76% reflection rate. The conversation highlights the importance of understanding units and the relationship between power, energy, and time in solving the problem.
Richie Smash
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Homework Statement


First I will say, I know I post here alot, 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.
 
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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.
 
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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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