#40

Am I missing something? We are getting the solar energy at the rate of 1 kW/m^2 but over how long of a time period? Is that per minute? Per hour?

Anyhow, let's say it's per hour. Then the 15-m^2 perfect panel will collect 15kW in one hour. So we need 40kW and divide by 15 to get 2 hours and 40 minutes.

Right?

no you are not missing anything but whoever wrote the question did
would support your assumption that it should read 1kWh/m^2 (or per hour somewhere else)

/oops

your post confused me a bit
there is nothing missing in the question since the rate converts to rate times hour over an hour

This post was edited by brmv on Oct 13 2013 06:43pm

I did get 2 2/3 hour also.

But isn't 1 watt 1 joule over a period of 1 second?l

so 1 kw would be 1000(1 joule) - 1000joules/sec

1000(3600) = 3.6 E6

That's all I know. I didn't know if the answer I got was right.

That part is more advanced than I know.

It's simply 2 hours and 40 minutes.

W = J/s but you don't need to do any conversions at all for this problem. It really is as simple as it seems.

40/15 = 2.67h

So, I was accidently right because I didn't know enough.

Thanks for explaining it.